✨Danh sách tích phân với phân thức

Sau đây là danh sách các tích phân (nguyên hàm) của các hàm phân thức. Tích phân của mọi hàm phân thức đều có thể được tính bằng phân tích phân số một phần thành tổng các hàm số có dạng: : $\backslash frac\{a\}\{(x-b)^n\}$, và $\backslash frac\{ax\; +\; b\}\{\backslash left((x-c)^2+d^2\backslash right)^n\}.$ rồi lần lượt xử lý từng số hạng.

Với những dạng hàm số khác, xem danh sách tích phân.

Hàm có dạng x^m(ax + b)ⁿ

Nhiều nguyên hàm dưới đây có hạng tử dạng . Do hạng tử này không có nghĩa khi , dạng tổng quát của nguyên hàm thay hằng số tích phân bằng một hàm hằng cục bộ. Tuy nhiên, người ta thường bỏ nó ra khỏi biểu thức. Ví dụ : thường được viết ngắn gọn là :$\backslash int\backslash frac\{1\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C,$ trong đó được hiểu là ký hiệu cho hàm hằng cục bộ ẩn . Quy ước này sẽ được tuân theo trong phần còn lại This convention will be adhered to in the following.

:$\backslash int\; (ax\; +\; b)^n\; \backslash ,\; dx=\; \backslash frac\{(ax\; +\; b)^\{n+1\{a(n\; +\; 1)\}\; +\; C\; \backslash qquad\; (n\backslash neq\; -1\backslash mbox\{)\}$ (Công thức diện tích Cavalieri) :$\backslash int\backslash frac\{x\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{x\}\{a\}\; -\; \backslash frac\{b\}\{a^2\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C$ :$\backslash int\backslash frac\{x\}\{(ax\; +\; b)^2\}\; \backslash ,\; dx=\; \backslash frac\{b\}\{a^2(ax\; +\; b)\}\; +\; \backslash frac\{1\}\{a^2\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C$ :$\backslash int\backslash frac\{x\}\{(ax\; +\; b)^n\}\; \backslash ,\; dx=\; \backslash frac\{a(1\; -\; n)x\; -\; b\}\{a^2(n\; -\; 1)(n\; -\; 2)(ax\; +\; b)^\{n-1\; +\; C\; \backslash qquad(n\backslash not\backslash in\; \{1,\; 2\}\backslash mbox\{)\}$ :$\backslash int\; x(ax\; +\; b)^n\; \backslash ,\; dx=\; \backslash frac\{a(n\; +\; 1)x\; -\; b\}\{a^2(n\; +\; 1)(n\; +\; 2)\}\; (ax\; +\; b)^\{n+1\}\; +\; C\; \backslash qquad(n\; \backslash not\backslash in\; \{-1,\; -2\}\backslash mbox\{)\}$ :$\backslash int\backslash frac\{x^2\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{b^2\backslash ln(\backslash left|ax\; +\; b\backslash right|)\}\{a^3\}+\backslash frac\{ax^2\; -\; 2bx\}\{2a^2\}\; +\; C$ :$\backslash int\backslash frac\{x^2\}\{(ax\; +\; b)^2\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a^3\}\backslash left(ax\; -\; 2b\backslash ln\backslash left|ax\; +\; b\backslash right|\; -\; \backslash frac\{b^2\}\{ax\; +\; b\}\backslash right)\; +\; C$ :$\backslash int\backslash frac\{x^2\}\{(ax\; +\; b)^3\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a^3\}\backslash left(\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; \backslash frac\{2b\}\{ax\; +\; b\}\; -\; \backslash frac\{b^2\}\{2(ax\; +\; b)^2\}\backslash right)\; +\; C$ :$\backslash int\backslash frac\{x^2\}\{(ax\; +\; b)^n\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a^3\}\backslash left(-\backslash frac\{(ax\; +\; b)^\{3-n\{(n-3)\}\; +\; \backslash frac\{2b\; (ax\; +\; b)^\{2-n\{(n-2)\}\; -\; \backslash frac\{b^2\; (ax\; +\; b)^\{1-n\{(n\; -\; 1)\}\backslash right)\; +\; C\; \backslash qquad(n\backslash not\backslash in\; \{1,\; 2,\; 3\}\backslash mbox\{)\}$ :$\backslash int\backslash frac\{1\}\{x(ax\; +\; b)\}\; \backslash ,\; dx\; =\; -\backslash frac\{1\}\{b\}\backslash ln\backslash left|\backslash frac\{ax+b\}\{x\}\backslash right|\; +\; C$ :$\backslash int\backslash frac\{1\}\{x^2(ax+b)\}\; \backslash ,\; dx\; =\; -\backslash frac\{1\}\{bx\}\; +\; \backslash frac\{a\}\{b^2\}\backslash ln\backslash left|\backslash frac\{ax+b\}\{x\}\backslash right|\; +\; C$ :$\backslash int\backslash frac\{1\}\{x^2(ax+b)^2\}\; \backslash ,\; dx\; =\; -a\backslash left(\backslash frac\{1\}\{b^2(ax+b)\}\; +\; \backslash frac\{1\}\{ab^2x\}\; -\; \backslash frac\{2\}\{b^3\}\backslash ln\backslash left|\backslash frac\{ax+b\}\{x\}\backslash right|\backslash right)\; +\; C$

Hàm có dạng x^m / (a x² + b x + c)ⁿ

Với :
: :$\backslash int\backslash frac\{x\}\{ax^2+bx+c\}\; \backslash ,\; dx\; =\; \backslash frac\{1\}\{2a\}\backslash ln\backslash left|ax^2+bx+c\backslash right|-\backslash frac\{b\}\{2a\}\backslash int\backslash frac\{dx\}\{ax^2+bx+c\}\; +\; C$ : : $\backslash int\backslash frac\{1\}\{(ax^2+bx+c)^n\}\; \backslash ,\; dx=\; \backslash frac\{2ax+b\}\{(n-1)(4ac-b^2)(ax^2+bx+c)^\{n-1+\backslash frac\{(2n-3)2a\}\{(n-1)(4ac-b^2)\}\backslash int\backslash frac\{1\}\{(ax^2+bx+c)^\{n-1\; \backslash ,\; dx\; +\; C$ : $\backslash int\backslash frac\{x\}\{(ax^2+bx+c)^n\}\; \backslash ,\; dx=\; -\backslash frac\{bx+2c\}\{(n-1)(4ac-b^2)(ax^2+bx+c)^\{n-1-\backslash frac\{b(2n-3)\}\{(n-1)(4ac-b^2)\}\backslash int\backslash frac\{1\}\{(ax^2+bx+c)^\{n-1\; \backslash ,\; dx\; +\; C$ : $\backslash int\backslash frac\{1\}\{x(ax^2+bx+c)\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{2c\}\backslash ln\backslash left|\backslash frac\{x^2\}\{ax^2+bx+c\}\backslash right|-\backslash frac\{b\}\{2c\}\backslash int\backslash frac\{1\}\{ax^2+bx+c\}\; \backslash ,\; dx\; +\; C$

Hàm có dạng x^m (a + b xⁿ)^p

• Những công thức sau hạ số mũ của hàm dưới dấu tích phân nhưng vẫn giữ nguyên dạng của chúng, do đó có thể được dùng nhiều lần để đưa số mũ và xuống 0.
• Những công thức hạ bậc này có thể dùng cho hàm có số mũ nguyên hoặc hữu tỉ.

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^p\; dx\; =\; \backslash frac\{x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^p\}\{m+n\backslash ,p+1\}\backslash ,+\backslash ,\; \backslash frac\{a\backslash ,n\backslash ,p\}\{m+n\backslash ,p+1\}\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p-1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^p\; dx\; =\; -\backslash frac\{x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\{a\backslash ,n\; (p+1)\}\backslash ,+\backslash ,\; \backslash frac\{m+n\; (p+1)+1\}\{a\backslash ,n\; (p+1)\}\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^p\; dx\; =\; \backslash frac\{x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^p\}\{m+1\}\backslash ,-\backslash ,\; \backslash frac\{b\backslash ,n\backslash ,p\}\{m+1\}\backslash int\; x^\{m+n\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p-1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^p\; dx\; =\; \backslash frac\{x^\{m-n+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\{b\backslash ,n\; (p+1)\}\backslash ,-\backslash ,\; \backslash frac\{m-n+1\}\{b\backslash ,n\; (p+1)\}\backslash int\; x^\{m-n\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^p\; dx\; =\; \backslash frac\{x^\{m-n+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\{b\; (m+n\backslash ,p+1)\}\backslash ,-\backslash ,\; \backslash frac\{a\; (m-n+1)\}\{b\; (m+n\backslash ,p+1)\}\backslash int\; x^\{m-n\}\backslash left(a+b\backslash ,x^n\backslash right)^pdx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n\backslash right)^p\; dx\; =\; \backslash frac\{x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\{a\; (m+1)\}\backslash ,-\backslash ,\; \backslash frac\{b\; (m+n\; (p+1)+1)\}\{a\; (m+1)\}\backslash int\; x^\{m+n\}\backslash left(a+b\backslash ,x^n\backslash right)^pdx$

Hàm có dạng (A + B x) (a + b x)^m (c + d x)ⁿ (e + f x)^p

• Tương tự như trên, những công thức hạ bậc này có thể được dùng nhiều lần để đưa , và xuống 0.
• Những công thức này dùng được cho số mũ là số nguyên hoặc số hữu tỉ.
• Cho bằng 0, ta có trường hợp đặc biệt $(a+b\backslash ,x)^m\; (c+d\backslash ,x)^n\; (e+f\backslash ,x)^p$.

:$\backslash int\; (A+B\backslash ,x)\; (a+b\backslash ,x)^m\; (c+d\backslash ,x)^n\; (e+f\backslash ,x)^p\; dx=\; -\backslash frac\{(A\backslash ,b-a\backslash ,B)(a+b\backslash ,x)^\{m+1\}\; (c+d\backslash ,x)^n(e+f\backslash ,x)^\{p+1\{b\; (m+1)\; (a\backslash ,f-b\backslash ,e)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{b\; (m+1)\; (a\backslash ,f-b\backslash ,e)\}\backslash ,\backslash cdot$

::$\backslash int\; (b\backslash ,c(m+1)\; (A\backslash ,f-B\backslash ,e)+(A\backslash ,b-a\backslash ,B)\; (n\backslash ,d\backslash ,e+c\backslash ,f(p+1))+d(b(m+1)\; (A\backslash ,f-B\backslash ,e)+f(n+p+1)\; (A\backslash ,b-a\backslash ,B))x)(a+b\backslash ,x)^\{m+1\}\; (c+d\backslash ,x)^\{n-1\}(e+f\backslash ,x)^p\; dx$

:$\backslash int\; (A+B\backslash ,x)\; (a+b\backslash ,x)^m\; (c+d\backslash ,x)^n\; (e+f\backslash ,x)^p\; dx=\; \backslash frac\{B(a+b\backslash ,x)^m\; (c+d\backslash ,x)^\{n+1\}(e+f\backslash ,x)^\{p+1\{d\backslash ,f(m+n+p+2)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{d\backslash ,f(m+n+p+2)\}\backslash ,\backslash cdot$

::$\backslash int\; (A\backslash ,a\backslash ,d\backslash ,f(m+n+p+2)-B\; (b\backslash ,c\backslash ,e\backslash ,m+a(d\backslash ,e(n+1)+c\backslash ,f(p+1)))+(A\backslash ,b\backslash ,d\backslash ,f(m+n+p+2)+B\; (a\backslash ,d\backslash ,f\backslash ,m-b(d\backslash ,e(m+n+1)+c\backslash ,f(m+p+1))))\; x)(a+b\backslash ,x)^\{m-1\}\; (c+d\backslash ,x)^n(e+f\backslash ,x)^p\; dx$

:$\backslash int\; (A+B\backslash ,x)\; (a+b\backslash ,x)^m\; (c+d\backslash ,x)^n\; (e+f\backslash ,x)^p\; dx=\; \backslash frac\{(A\backslash ,b-a\backslash ,B)(a+b\backslash ,x)^\{m+1\}\; (c+d\backslash ,x)^\{n+1\}(e+f\backslash ,x)^\{p+1\{(m+1)(a\backslash ,d-b\backslash ,c)(a\backslash ,f-b\backslash ,e)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{(m+1)(a\backslash ,d-b\backslash ,c)(a\backslash ,f-b\backslash ,e)\}\backslash ,\backslash cdot$

::$\backslash int\; ((m+1)\; (A\; (a\backslash ,d\backslash ,f-b(c\backslash ,f+d\backslash ,e))+B\backslash ,b\backslash ,c\backslash ,e)-(A\backslash ,b-a\backslash ,B)\; (d\backslash ,e(n+1)+c\backslash ,f(p+1))-d\backslash ,f(m+n+p+3)\; (A\backslash ,b-a\backslash ,B)x)(a+b\backslash ,x)^\{m+1\}\; (c+d\backslash ,x)^n(e+f\backslash ,x)^p\; dx$

Hàm có dạng x^m (A + B xⁿ) (a + b xⁿ)^p (c + d xⁿ)^q

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; -\backslash frac\{(A\backslash ,b-a\backslash ,B)\; x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^q\}\{a\backslash ,b\backslash ,n\; (p+1)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{a\backslash ,b\backslash ,n\; (p+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^m\backslash left(c\; (A\backslash ,b\backslash ,n\; (p+1)+(A\backslash ,b-a\backslash ,B)\; (m+1))+d\; (A\backslash ,b\backslash ,n\; (p+1)+(A\backslash ,b-a\backslash ,B)\; (m+n\backslash ,q+1))\; x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\backslash left(c+d\backslash ,x^n\backslash right)^\{q-1\}dx$

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; \backslash frac\{B\backslash ,x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^q\}\{b\; (m+n\; (p+q+1)+1)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{b\; (m+n\; (p+q+1)+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^m\backslash left(c\; ((A\backslash ,b-a\backslash ,B)\; (1+m)+A\backslash ,b\backslash ,n\; (1+p+q))+(d(A\backslash ,b-a\backslash ,B)\; (1+m)+B\backslash ,n\backslash ,q(b\backslash ,c-a\backslash ,d)+A\backslash ,b\backslash ,d\backslash ,n\; (1+p+q))\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^\{q-1\}dx$

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; -\backslash frac\{(A\backslash ,b-a\backslash ,B)\; x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^\{q+1\{a\backslash ,n\; (b\backslash ,c-a\backslash ,d)\; (p+1)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{a\backslash ,n(b\backslash ,c-a\backslash ,d)(p+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^m\backslash left(c(A\backslash ,b-a\backslash ,B)(m+1)+A\backslash ,n\; (b\backslash ,c-a\backslash ,d)(p+1)+d(A\backslash ,b-a\backslash ,B)\; (m+n\; (p+q+2)+1)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\backslash left(c+d\backslash ,x^n\backslash right)^qdx$

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; \backslash frac\{B\backslash ,x^\{m-n+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^\{q+1\{b\backslash ,d\; (m+n\; (p+q+1)+1)\}\backslash ,-\backslash ,\; \backslash frac\{1\}\{b\backslash ,d\; (m+n\; (p+q+1)+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m-n\}\backslash left(a\backslash ,B\backslash ,c\; (m-n+1)+(a\backslash ,B\backslash ,d\; (m+n\backslash ,q+1)-b\; (-B\backslash ,c\; (m+n\backslash ,p+1)+A\backslash ,d\; (m+n\; (p+q+1)+1)))\; x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx$

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; \backslash frac\{A\backslash ,x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^\{q+1\{a\backslash ,c\; (m+1)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{a\backslash ,c\; (m+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m+n\}\backslash left(a\backslash ,B\backslash ,c\; (m+1)-A\; (b\backslash ,c+a\backslash ,d)\; (m+n+1)-A\backslash ,n\; (b\backslash ,c\backslash ,p+a\backslash ,d\backslash ,q)-A\backslash ,b\backslash ,d\; (m+n\; (p+q+2)+1)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx$

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; \backslash frac\{A\backslash ,x^\{m+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^q\}\{a\; (m+1)\}\backslash ,-\backslash ,\; \backslash frac\{1\}\{a\; (m+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m+n\}\backslash left(c(A\backslash ,b-a\backslash ,B)(m+1)+A\backslash ,n\; (b\backslash ,c\; (p+1)+a\backslash ,d\backslash ,q)+d\; ((A\backslash ,b-a\backslash ,B)\; (m+1)+A\backslash ,b\backslash ,n\; (p+q+1))\; x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^\{q-1\}dx$

:$\backslash int\; x^m\backslash left(A+B\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^qdx=\; \backslash frac\{(A\backslash ,b-a\backslash ,B)\; x^\{m-n+1\}\; \backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\; \backslash left(c+d\backslash ,x^n\backslash right)^\{q+1\{b\backslash ,n\; (b\backslash ,c-a\backslash ,d)\; (p+1)\}\backslash ,-\backslash ,\; \backslash frac\{1\}\{b\backslash ,n(b\backslash ,c-a\backslash ,d)(p+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m-n\}\backslash left(c(A\backslash ,b-a\backslash ,B)(m-n+1)+(d(A\backslash ,b-a\backslash ,B)(m+n\backslash ,q+1)-b\backslash ,n(B\backslash ,c-A\backslash ,d)(p+1))\; x^n\backslash right)\backslash left(a+b\backslash ,x^n\backslash right)^\{p+1\}\backslash left(c+d\backslash ,x^n\backslash right)^qdx$

Hàm có dạng (d + e x)^m (a + b x + c x²)^p với b² − 4 a c = 0

:$\backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{e(m+1)\}\backslash ,-\backslash ,\; \backslash frac\{p\; (d+e\backslash ,x)^\{m+2\}(b+2\; c\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\{e^2(m+1)(m+2\; p+1)\}\backslash ,+\backslash ,\; \backslash frac\{p(2\; p-1)(2\; c\backslash ,d-b\backslash ,e)\}\{e^2(m+1)(m+2\; p+1)\}\; \backslash int\; (d+e\backslash ,x)^\{m+1\}\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{e(m+1)\}\backslash ,-\backslash ,\; \backslash frac\{p\; (d+e\backslash ,x)^\{m+2\}(b+2\backslash ,c\backslash ,x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\{e^2(m+1)(m+2)\}\backslash ,+\backslash ,\; \backslash frac\{2\backslash ,c\backslash ,p\backslash ,(2\backslash ,p-1)\}\{e^2(m+1)(m+2)\}\; \backslash int\; (d+e\backslash ,x)^\{m+2\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; -\backslash frac\{e(m+2\; p+2)(d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{(p+1)(2p+1)(2\; c\backslash ,d-b\backslash ,e)\}\backslash ,+\backslash ,\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}(b+2\; c\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{(2p+1)(2\; c\backslash ,d-b\backslash ,e)\}\backslash ,+\backslash ,\; \backslash frac\{e^2m(m+2\; p+2)\}\{(p+1)(2p+1)(2\; c\backslash ,d-b\backslash ,e)\}\; \backslash int\; (d+e\backslash ,x)^\{m-1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; -\backslash frac\{e\backslash ,m(d+e\backslash ,x)^\{m-1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{2c\; (p+1)\; (2p+1)\}\backslash ,+\backslash ,\; \backslash frac\{(d+e\backslash ,x)^m(b+2\; c\backslash ,x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{2c\; (2p+1)\}\backslash ,+\backslash ,\; \backslash frac\{e^2m(m-1)\}\{2c\; (p+1)\; (2p+1)\}\; \backslash int\; (d+e\backslash ,x)^\{m-2\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{e(m+2p+1)\}\backslash ,-\backslash ,\; \backslash frac\{p(2\; c\backslash ,d-b\backslash ,e)(d+e\backslash ,x)^\{m+1\}(b+2\; c\backslash ,x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\{2c\backslash ,e^2(m+2\; p)(m+2p+1)\}\backslash ,+\backslash ,\; \backslash frac\{p\; (2\; p-1)(2\; c\backslash ,d-b\backslash ,e)^2\}\{2c\backslash ,e^2(m+2\; p)(m+2p+1)\}\; \backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; -\backslash frac\{2c\backslash ,e(m+2p+2)(d+e\backslash ,x)^\{m+1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{(p+1)\; (2\; p+1)(2\; c\backslash ,d-b\backslash ,e)^2\}\backslash ,+\backslash ,\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}(b+2\; c\backslash ,x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{(2\; p+1)(2\; c\backslash ,d-b\backslash ,e)\}\backslash ,+\backslash ,\; \backslash frac\{2c\backslash ,e^2(m+2p+2)(m+2\; p+3)\}\{(p+1)\; (2\; p+1)(2\; c\backslash ,d-b\backslash ,e)^2\}\; \backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^m\; (b+2\; c\backslash ,x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{2c\; (m+2p+1)\}\backslash ,+\backslash ,\; \backslash frac\{m(2\; c\backslash ,d-b\backslash ,e)\}\{2c\; (m+2p+1)\}\; \backslash int\; (d+e\backslash ,x)^\{m-1\}\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx$

:$\backslash int\; (d+e\backslash ,x)^m\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; -\backslash frac\{(d+e\backslash ,x)^\{m+1\}\; (b+2\; c\backslash ,x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{(m+1)(2\; c\backslash ,d-b\backslash ,e)\}\backslash ,+\backslash ,\; \backslash frac\{2c\; (m+2p+2)\}\{(m+1)(2\; c\backslash ,d-b\backslash ,e)\}\; \backslash int\; (d+e\backslash ,x)^\{m+1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx$

Hàm có dạng (d + e x)^m (A + B x) (a + b x + c x²)^p

:$\backslash int\; (d+e\backslash ,x)^m\; (A+B\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}\; (A\backslash ,e\; (m+2\; p+2)-B\backslash ,d\; (2\; p+1)+e\backslash ,B\; (m+1)\; x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{e^2(m+1)\; (m+2\; p+2)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{e^2(m+1)\; (m+2\; p+2)\}p\backslash ,\backslash cdot$

::$\backslash int\; (d+e\backslash ,x)^\{m+1\}\; (B\; (b\backslash ,d+2\; a\backslash ,e+2\; a\backslash ,e\backslash ,m+2\; b\backslash ,d\backslash ,p)-A\backslash ,b\backslash ,e\; (m+2\; p+2)+(B\; (2\; c\backslash ,d+b\backslash ,e+b\backslash ,e\; m+4\; c\backslash ,d\backslash ,p)-2\; A\backslash ,c\backslash ,e\; (m+2\; p+2))x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; (A+B\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^m\; (A\backslash ,b-2\; a\backslash ,B-(b\backslash ,B-2\; A\backslash ,c)\; x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{(p+1)\backslash left(b^2-4\; a\backslash ,c\backslash right)\; \}\backslash ,+\backslash ,\; \backslash frac\{1\}\{(p+1)\backslash left(b^2-4\; a\backslash ,c\backslash right)\; \}\backslash ,\backslash cdot$

::$\backslash int\; (d+e\backslash ,x)^\{m-1\}(B\; (2\; a\backslash ,e\backslash ,m+b\backslash ,d\; (2\; p+3))-A\; (b\backslash ,e\backslash ,m+2\; c\backslash ,d\; (2\; p+3))+e(b\backslash ,B-2\; A\backslash ,c)\; (m+2\; p+3)\; x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; (A+B\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}\; (A\backslash ,c\backslash ,e\; (m+2\; p+2)-B\; (c\backslash ,d+2\; c\backslash ,d\backslash ,p-b\backslash ,e\backslash ,p)+B\backslash ,c\backslash ,e(m+2\; p+1)\; x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p\}\{c\backslash ,e^2(m+2\; p+1)\; (m+2\; p+2)\}\backslash ,-\backslash ,\; \backslash frac\{p\}\{c\backslash ,e^2(m+2\; p+1)\; (m+2\; p+2)\}\backslash ,\backslash cdot$

::$\backslash int\; (d+e\backslash ,x)^m\; (A\backslash ,c\backslash ,e\; (b\backslash ,d-2\; a\backslash ,e)\; (m+2\; p+2)+B\; (a\backslash ,e\; (b\backslash ,e-2\; c\backslash ,d\backslash ,m+b\backslash ,e\backslash ,m)+b\backslash ,d\; (b\backslash ,e\backslash ,p-c\backslash ,d-2\; c\backslash ,d\backslash ,p))+$

:::$\backslash left(A\backslash ,c\backslash ,e\; (2\; c\backslash ,d-b\backslash ,e)\; (m+2\; p+2)-B\; \backslash left(-b^2\; e^2\; (m+p+1)+2\; c^2\; d^2\; (1+2\; p)+c\backslash ,e\; (b\backslash ,d\; (m-2\; p)+2\; a\backslash ,e\; (m+2\; p+1))\backslash right)\backslash right)\; x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p-1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; (A+B\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{(d+e\backslash ,x)^\{m+1\}\; \backslash left(A\; \backslash left(b\backslash ,c\backslash ,d-b^2\; e+2\; a\backslash ,c\backslash ,e\backslash right)-a\backslash ,B\; (2\; c\backslash ,d-b\backslash ,e)+c\; (A\; (2\; c\backslash ,d-b\backslash ,e)-B\; (b\backslash ,d-2\; a\backslash ,e))\; x\backslash right)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{(p+1)\backslash left(b^2-4\; a\backslash ,c\backslash right)\; \backslash left(c\backslash ,d^2-b\backslash ,d\backslash ,e+a\backslash ,e^2\backslash right)\}\backslash ,+$

::$\backslash frac\{1\}\{(p+1)\backslash left(b^2-4\; a\backslash ,c\backslash right)\; \backslash left(c\backslash ,d^2-b\backslash ,d\backslash ,e+a\backslash ,e^2\backslash right)\}\backslash ,\backslash cdot$

:::$\backslash int\; (d+e\backslash ,x)^m\; (A\; \backslash left(b\backslash ,c\backslash ,d\backslash ,e\; (2\; p-m+2)+b^2\; e^2\; (m+p+2)-2\; c^2\; d^2\; (3+2\; p)-2\; a\backslash ,c\backslash ,e^2\; (m+2\; p+3)\backslash right)-$

::::$B\; (a\backslash ,e\; (b\backslash ,e-2\; c\backslash ,d\; m+b\backslash ,e\backslash ,m)+b\backslash ,d\; (-3\; c\backslash ,d+b\backslash ,e-2\; c\backslash ,d\backslash ,p+b\backslash ,e\backslash ,p))+c\backslash ,e(B\; (b\backslash ,d-2\; a\backslash ,e)-A\; (2\; c\backslash ,d-b\backslash ,e))\; (m+2\; p+4)\; x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\}dx$

:$\backslash int\; (d+e\backslash ,x)^m\; (A+B\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; \backslash frac\{B(d+e\backslash ,x)^m\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{c(m+2\; p+2)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{c(m+2\; p+2)\}\backslash ,\backslash cdot$

::$\backslash int\; (d+e\backslash ,x)^\{m-1\}\; (m(A\backslash ,c\backslash ,d-a\backslash ,B\backslash ,e)-d(b\backslash ,B-2\; A\backslash ,c)(p+1)\; +((B\backslash ,c\backslash ,d-b\backslash ,B\backslash ,e+A\backslash ,c\backslash ,e)\; m-e(b\backslash ,B-2\; A\backslash ,c)(p+1))x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx$

:$\backslash int\; (d+e\backslash ,x)^m\; (A+B\backslash ,x)\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx=\; -\backslash frac\{(B\backslash ,d-A\backslash ,e)\; (d+e\backslash ,x)^\{m+1\}\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^\{p+1\{(m+1)\backslash left(c\backslash ,d^2-b\backslash ,d\backslash ,e+a\backslash ,e^2\backslash right)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{(m+1)\backslash left(c\backslash ,d^2-b\backslash ,d\backslash ,e+a\backslash ,e^2\backslash right)\}\backslash ,\backslash cdot$

::$\backslash int\; (d+e\backslash ,x)^\{m+1\}\; ((A\backslash ,c\backslash ,d-A\backslash ,b\backslash ,e+a\backslash ,B\backslash ,e)\; (m+1)+b\; (B\backslash ,d-A\backslash ,e)\; (p+1)+c\; (B\backslash ,d-A\backslash ,e)\; (m+2\; p+3)\; x)\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^pdx$

== Hàm có dạng x^m (a + b xⁿ + c x²ⁿ)^p với {(m+1)(m+2 n\,p+1)}\,-\, \frac{b\,n^2 p (2 p-1)}{(m+1)(m+2 n\,p+1)} \int x^{m+n} \left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}dx

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; \backslash frac\{(m+n(2\; p-1)+1)\; x^\{m+1\}\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{(m+1)(m+n+1)\}\backslash ,+\backslash ,\; \backslash frac\{n\backslash ,p\backslash ,x^\{m+1\}\; \backslash left(2\; a+b\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p-1\{(m+1)(m+n+1)\}\backslash ,+\backslash ,\; \backslash frac\{2\; c\backslash ,p\backslash ,n^2(2\; p-1)\}\{(m+1)(m+n+1)\}\; \backslash int\; x^\{m+2n\}\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p-1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; \backslash frac\{(m+n(2\; p+1)+1)\; x^\{m-n+1\}\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{b\backslash ,n^2\; (p+1)\; (2p+1)\}\backslash ,-\backslash ,\; \backslash frac\{x^\{m+1\}\; \backslash left(b+2\; c\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{b\backslash ,n\; (2p+1)\}\backslash ,-\backslash ,\; \backslash frac\{(m-n+1)(m+n(2\; p+1)+1)\}\{b\backslash ,n^2\; (p+1)\; (2p+1)\}\; \backslash int\; x^\{m-n\}\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; -\backslash frac\{(m-3\; n-2\; n\backslash ,p+1)\; x^\{m-2n+1\}\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{2\; c\backslash ,n^2(p+1)(2p+1)\}\backslash ,-\backslash ,\; \backslash frac\{\; x^\{m-2n+1\}\; \backslash left(2\; a+b\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{2\; c\backslash ,n(2p+1)\}\backslash ,+\backslash ,\; \backslash frac\{(m-n+1)(m-2n+1)\}\{2\; c\backslash ,n^2(p+1)(2p+1)\}\; \backslash int\; x^\{m-2n\}\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; \backslash frac\{x^\{m+1\}\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{m+2\; n\backslash ,p+1\}\backslash ,+\backslash ,\; \backslash frac\{n\backslash ,p\backslash ,x^\{m+1\}\; \backslash left(2\; a+b\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p-1\{(m+2\; n\backslash ,p+1)\; (m+n(2\; p-1)+1)\}\backslash ,+\backslash ,\; \backslash frac\{2\; a\backslash ,n^2\; p\; (2\; p-1)\}\{(m+2\; n\backslash ,p+1)\; (m+n(2\; p-1)+1)\}\; \backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p-1\}dx$

:$\backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; -\backslash frac\{(m+n+2\; n\backslash ,p+1)\; x^\{m+1\}\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{2\; a\backslash ,n^2\; (p+1)\; (2p+1)\}\backslash ,-\backslash ,\; \backslash frac\{x^\{m+1\}\; \backslash left(2\; a+b\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{2\; a\backslash ,n(2p+1)\}\backslash ,+\backslash ,\; \backslash frac\{(m+n(2\; p+1)+1)(m+2\; n\; (p+1)+1)\}\{2\; a\backslash ,n^2\; (p+1)\; (2p+1)\}\; \backslash int\; x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; \backslash frac\{x^\{m-n+1\}\; \backslash left(b+2c\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{2c\; (m+2n\backslash ,p+1)\}\backslash ,-\backslash ,\; \backslash frac\{b\; (m-n+1)\}\{2c\; (m+2n\backslash ,p+1)\}\; \backslash int\; x^\{m-n\}\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx$

:$\backslash int\; x^m\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx=\; \backslash frac\{x^\{m+1\}\; \backslash left(b+2c\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{b\; (m+1)\}\backslash ,-\backslash ,\; \backslash frac\{2c\; (m+n(2\; p+1)+1)\}\{b\; (m+1)\}\; \backslash int\; x^\{m+n\}\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\; dx$

Hàm có dạng x^m (A + B xⁿ) (a + b xⁿ + c x²ⁿ)^p

:$\backslash int\; x^m\; \backslash left(A+B\backslash ,x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx=\; \backslash frac\{x^\{m+1\}\; \backslash left(A\; (m+n\; (2\; p+1)+1)+B\; (m+1)\; x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{(m+1)\; (m+n\; (2\; p+1)+1)\}\backslash ,+\backslash ,\; \backslash frac\{n\backslash ,p\}\{(m+1)\; (m+n\; (2\; p+1)+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m+n\}\; \backslash left(2\; a\backslash ,B\; (m+1)-A\backslash ,b\; (m+n\; (2\; p+1)+1)+(b\backslash ,B\; (m+1)-2\backslash ,A\backslash ,c\; (m+n\; (2\; p+1)+1))\; x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p-1\}dx$

:$\backslash int\; x^m\; \backslash left(A+B\backslash ,x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx=\; \backslash frac\{x^\{m-n+1\}\; \backslash left(A\backslash ,b-2\; a\backslash ,B-(b\backslash ,B-2\; A\backslash ,c)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{n(p+1)\; \backslash left(b^2-4\; a\backslash ,c\backslash right)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{n(p+1)\; \backslash left(b^2-4\; a\backslash ,c\backslash right)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m-n\}\backslash left((m-n+1)(2\; a\backslash ,B-A\backslash ,b)+(m+2n\; (p+1)+1)\; (b\backslash ,B-2\; A\backslash ,c)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\; \backslash left(A+B\backslash ,x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx=\; \backslash frac\{x^\{m+1\}\; \backslash left(b\backslash ,B\backslash ,n\backslash ,p+A\backslash ,c\; (m+n\; (2\; p+1)+1)+B\backslash ,c\; (m+2\; n\backslash ,p+1)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p\}\{c\; (m+2\; n\backslash ,p+1)\; (m+n\; (2\; p+1)+1)\}\backslash ,+\backslash ,\; \backslash frac\{n\backslash ,p\}\{c\; (m+2\; n\backslash ,p+1)\; (m+n\; (2\; p+1)+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^m\; \backslash left(2\; a\backslash ,A\backslash ,c\; (m+n\; (2\; p+1)+1)-a\backslash ,b\backslash ,B\; (m+1)+\backslash left(2\; a\backslash ,B\backslash ,c\; (m+2\; n\backslash ,p+1)+A\backslash ,b\backslash ,c\; (m+n\; (2\; p+1)+1)-b^2\; B\; (m+n\backslash ,p+1)\backslash right)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p-1\}dx$

:$\backslash int\; x^m\; \backslash left(A+B\backslash ,x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx=\; -\backslash frac\{x^\{m+1\}\; \backslash left(A\backslash ,b^2-a\backslash ,b\backslash ,B-2\; a\backslash ,A\backslash ,c+(A\backslash ,b-2\; a\backslash ,B)\; c\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{a\backslash ,n(p+1)\; \backslash left(b^2-4\; a\backslash ,c\backslash right)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{a\backslash ,n(p+1)\; \backslash left(b^2-4\; a\backslash ,c\backslash right)\}\backslash ,\backslash cdot$

::$\backslash int\; x^m\; \backslash left((m+n\; (p+1)+1)\; A\backslash ,b^2-a\backslash ,b\backslash ,B(m+1)-2(m+2n\; (p+1)+1)a\backslash ,A\backslash ,c+(m+n\; (2p+3)+1)(A\backslash ,b-2\; a\backslash ,B)\; c\backslash ,x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\}dx$

:$\backslash int\; x^m\; \backslash left(A+B\backslash ,x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx=\; \backslash frac\{B\backslash ,x^\{m-n+1\}\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{c\; (m+n\; (2\; p+1)+1)\}\backslash ,-\backslash ,\; \backslash frac\{1\}\{c\; (m+n\; (2\; p+1)+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m-n\}\; \backslash left(a\backslash ,B\; (m-n+1)+(b\backslash ,B\; (m+n\backslash ,p+1)-A\backslash ,c\; (m+n\; (2\; p+1)+1))\; x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx$

:$\backslash int\; x^m\; \backslash left(A+B\backslash ,x^n\backslash right)\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx=\; \backslash frac\{A\backslash ,x^\{m+1\}\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^\{p+1\{a(m+1)\}\backslash ,+\backslash ,\; \backslash frac\{1\}\{a(m+1)\}\backslash ,\backslash cdot$

::$\backslash int\; x^\{m+n\}\; \backslash left(a\backslash ,B\; (m+1)-A\backslash ,b\; (m+n\; (p+1)+1)-A\backslash ,c\; (m+2\; n(p+1)+1)\; x^n\backslash right)\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^pdx$

Các hàm khác

:$\backslash int\backslash frac\{f\text{'}(x)\}\{f(x)\}\; \backslash ,\; dx=\; \backslash ln\backslash left|f(x)\backslash right|\; +\; C$

:$\backslash int\backslash frac\{1\}\{x^2+a^2\}\; \backslash ,\; dx\; =\; \backslash frac\{1\}\{a\}\backslash arctan\backslash frac\{x\}\{a\}\backslash ,!\; +\; C$ : :$\backslash int\; \backslash frac\{dx\}\{x^\{2^n\}\; +\; 1\}\; =\; \backslash frac\{1\}\{2^\{n-1\backslash sum\_\{k=1\}^\{2^\{n-1\; \backslash sin\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\backslash right)\; \backslash arctan\backslash left[\backslash left(x\; -\; \backslash cos\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; \backslash right)\; \backslash csc\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; \backslash right]\; -\; \backslash frac\{1\}\{2\}\; \backslash cos\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; \backslash ln\; \backslash left\; |\; x^2\; -\; 2\; x\; \backslash cos\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; +\; 1\; \backslash right\; |\; +\; C$