List of integrals of rational functions

The following is a list of integrals (antiderivative functions) of rational functions. For a more complete list of integrals, see lists of integrals.

$\int (ax + b)^n \, dx= \frac{(ax + b)^{n+1}}{a(n + 1)} + C \qquad\text{(for } n\neq -1\mbox{)}\,\!$ (Cavalieri's quadrature formula)

$\int\frac{c}{ax + b} \, dx= \frac{c}{a}\ln\left|ax + b\right| + C$

$\int x(ax + b)^n \, dx= \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} + C \qquad\text{(for }n \not\in \{-1, -2\}\mbox{)}$

$\int\frac{x}{ax + b} \, dx= \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right| + C$

$\int\frac{x}{(ax + b)^2} \, dx= \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right| + C$

$\int\frac{x}{(ax + b)^n} \, dx= \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}} + C \qquad\text{(for } n\not\in \{1, 2\}\mbox{)}$

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

$\int\frac{x^2}{ax + b} \, dx= \frac{b^2\ln(\left|ax + b\right|)}{a^3}+\frac{ax^2 - 2bx}{2a^2} + C$

$\int\frac{x^2}{(ax + b)^2} \, dx= \frac{1}{a^3}\left(ax - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right) + C$

$\int\frac{x^2}{(ax + b)^3} \, dx= \frac{1}{a^3}\left(\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right) + C$

$\int\frac{x^2}{(ax + b)^n} \, dx= \frac{1}{a^3}\left(-\frac{(ax + b)^{3-n}}{(n-3)} + \frac{2b (ax + b)^{2-n}}{(n-2)} - \frac{b^2 (ax + b)^{1-n}}{(n - 1)}\right) + C \qquad\text{(for } n\not\in \{1, 2, 3\}\mbox{)}$

$\int\frac{1}{x(ax + b)} \, dx = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right| + C$

$\int\frac{1}{x^2(ax+b)} \, dx = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right| + C$

$\int\frac{1}{x^2(ax+b)^2} \, dx = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right) + C$

$\int\frac{1}{x^2+a^2} \, dx = \frac{1}{a}\arctan\frac{x}{a}\,\! + C$

$\int\frac{1}{x^2-a^2} \, dx = \begin{cases} \displaystyle -\frac{1}{a}\,\mathrm{arctanh}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x} + C & \text{(for }|x| < |a|\mbox{)} \\[12pt] \displaystyle -\frac{1}{a}\,\mathrm{arccoth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a} + C & \text{(for }|x| > |a| \mbox{)} \end{cases}$

$\int \frac{dx}{x^{2^n} + 1} = \sum_{k=1}^{2^{n-1}} \left \{ \frac{1}{2^{n-1}} \left [ \sin \left(\frac{(2k -1) \pi}{2^n}\right) \arctan\left[\left(x - \cos \left(\frac{(2k -1) \pi}{2^n} \right) \right ) \csc \left(\frac{(2k -1) \pi}{2^n} \right) \right] \right] - \frac{1}{2^n} \left [ \cos \left(\frac{(2k -1) \pi}{2^n} \right) \ln \left | x^2 - 2 x \cos \left(\frac{(2k -1) \pi}{2^n} \right) + 1 \right | \right ] \right \} + C$

Any rational function can be integrated using the above equations and partial fractions in integration, by decomposing the rational function into a sum of functions of the form:

$\frac{a}{(x-b)^n}$, and $\frac{ax + b}{\left((x-c)^2+d^2\right)^n}.$

Integrands of the form $\frac{x^m}{(a\,x^2+b\,x+c)^n}$

For $a\neq 0:$

$\int\frac{1}{ax^2+bx+c} dx = \begin{cases} \displaystyle \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C & \text{(for }4ac-b^2>0\mbox{)} \\[12pt] \displaystyle -\frac{2}{\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C = \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C & \text{(for }4ac-b^2<0\mbox{)} \\[12pt] \displaystyle -\frac{2}{2ax+b} + C & \text{(for }4ac-b^2=0\mbox{)} \end{cases}$

$\int\frac{x}{ax^2+bx+c} \, dx = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c} + C$

$\int\frac{mx+n}{ax^2+bx+c} \, dx = \begin{cases} \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C &\text{(for }4ac-b^2>0\mbox{)} \\[12pt] \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C &\text{(for }4ac-b^2<0\mbox{)} \\[12pt] \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} + C &\text{(for }4ac-b^2=0\mbox{)}\end{cases}$

$\int\frac{1}{(ax^2+bx+c)^n} \, dx= \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int\frac{1}{(ax^2+bx+c)^{n-1}} \, dx + C$

$\int\frac{x}{(ax^2+bx+c)^n} \, dx= -\frac{bx+2c}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int\frac{1}{(ax^2+bx+c)^{n-1}} \, dx + C$

$\int\frac{1}{x(ax^2+bx+c)} \, dx= \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right|-\frac{b}{2c}\int\frac{1}{ax^2+bx+c} \, dx + C$

Integrands of the form $x^m \left(a+b\,x^n\right)^p$

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.

• These reduction formulas can be used for integrands having integer and/or fractional exponents.

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

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

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

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

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

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

Integrands of the form $x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^q$

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, p and q toward 0.

• These reduction formulas can be used for integrands having integer and/or fractional exponents.

• Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^q$ and $x^m\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^q$ by setting m and/or B to 0.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Integrands of the form $x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^p$

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.

• These reduction formulas can be used for integrands having integer and/or fractional exponents.

• Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x^n+c\,x^{2 n}\right)^p$ and $x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p$ by setting m and/or B to 0.

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

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

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

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

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

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

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

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

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

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

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

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