List of integrals of rational functions

The following is a list of integrals (antiderivative functions) of rational functions.
Any rational function can be integrated by partial fraction decomposition of the function into a sum of functions of the form:
: $\frac$, and $\frac.$
which can then be integrated term by term.

For other types of functions, see lists of integrals.

Miscellaneous integrands

:$\int\frac \, dx= \ln\left, f(x)\ + C$

:$\int\frac \, dx = \frac\arctan\frac\,\! + C$
:$\int\frac \, dx = \frac\ln\left, \frac\ + C = \begin \displaystyle -\frac\,\operatorname\frac + C = \frac\ln\frac + C & \text, x, < , a, \mbox \\$2pt\displaystyle -\frac\,\operatorname\frac + C = \frac\ln\frac + C & \text, x, > , a, \mbox \end
:$\int\frac \, dx = \frac\ln\left, \frac\ + C = \begin \displaystyle \frac\,\operatorname\frac + C = \frac\ln\frac + C & \text, x, < , a, \mbox \\$2pt\displaystyle \frac\,\operatorname\frac + C = \frac\ln\frac + C & \text, x, > , a, \mbox \end
: \int \frac = \frac\sum_^ \sin \left(\frac\pi\right) \arctan\left left(x - \cos \left(\frac\pi \right) \right ) \csc \left(\frac\pi \right) \right- \frac \cos \left(\frac\pi \right) \ln \left , x^2 - 2 x \cos \left(\frac\pi \right) + 1 \right , + C

Integrands of the form ''x''^''m''(''a x'' + ''b'')^''n''

Many of the following antiderivatives have a term of the form ln , ''ax'' + ''b'', . Because this is undefined when ''x'' = −''b'' / ''a'', the most general form of the antiderivative replaces the constant of integration with a locally constant function.Reader Survey: log, ''x'', + ''C''
, Tom Leinster, ''The ''n''-category Café'', March 19, 2012 However, it is conventional to omit this from the notation. For example,
:$\int\frac \, dx= \begin \dfrac\ln(-(ax + b)) + C^- & ax+b<0 \\ \dfrac\ln(ax + b) + C^+ & ax+b>0 \end$
is usually abbreviated as
:$\int\frac \, dx= \frac\ln\left, ax + b\ + C,$
where ''C'' is to be understood as notation for a locally constant function of ''x''. This convention will be adhered to in the following.

:$\int (ax + b)^n \, dx= \frac + C \qquad\text n\neq -1\mbox$ ( Cavalieri's quadrature formula)
:$\int\frac \, dx= \frac - \frac\ln\left, ax + b\ + C$
:$\int\frac \, dx= \frac x + \frac\ln\left, ax + b\ + C$
:$\int\frac \, dx= \frac + \frac\ln\left, ax + b\ + C$
:$\int\frac \, dx= \frac + C \qquad\text n\not\in \\mbox$
:$\int x(ax + b)^n \, dx= \frac (ax + b)^ + C \qquad\textn \not\in \\mbox$
:$\int\frac \, dx= \frac+\frac + C$
:$\int\frac \, dx= \frac\left(ax - 2b\ln\left, ax + b\ - \frac\right) + C$
:$\int\frac \, dx= \frac\left(\ln\left, ax + b\ + \frac - \frac\right) + C$
:$\int\frac \, dx= \frac\left(-\frac + \frac - \frac\right) + C \qquad\text n\not\in \\mbox$
:$\int\frac \, dx = -\frac\ln\left, \frac\ + C$
:$\int\frac \, dx = -\frac + \frac\ln\left, \frac\ + C$
:$\int\frac \, dx = -a\left(\frac + \frac - \frac\ln\left, \frac\\right) + C$

Integrands of the form ''x''^''m'' / (''a x''² + ''b x'' + ''c'')^''n''

For $a\neq 0:$


:$\int\frac dx = \begin \displaystyle \frac\arctan\frac + C & \text4ac-b^2>0\mbox \\$2pt\displaystyle \frac\ln\left, \frac\ + C =
\begin
\displaystyle -\frac\,\operatorname\frac + C &\text, 2ax+b, <\sqrt\mbox \\ pt\displaystyle -\frac\,\operatorname\frac + C &\text
\end
& \text4ac-b^2<0\mbox \\2pt\displaystyle -\frac + C & \text4ac-b^2=0\mbox
\end
:$\int\frac \, dx = \frac\ln\left, ax^2+bx+c\-\frac\int\frac + C$
:$\int\frac \, dx = \begin \displaystyle \frac\ln\left, ax^2+bx+c\+\frac\arctan\frac + C &\text4ac-b^2>0\mbox \\$2pt\displaystyle \frac\ln\left, ax^2+bx+c\+\frac\ln\left, \frac\ + C =
\begin
\displaystyle \frac\ln\left, ax^2+bx+c\-\frac\,\operatorname\frac + C &\text, 2ax+b, <\sqrt\mbox \\ pt\displaystyle \frac\ln\left, ax^2+bx+c\-\frac\,\operatorname\frac + C &\text
\end
& \text4ac-b^2<0\mbox \\2pt\displaystyle \frac\ln\left, ax^2+bx+c\-\frac + C = \frac\ln\left, x+\frac\-\frac + C &\text4ac-b^2=0\mbox\end
: $\int\frac \, dx= \frac+\frac\int\frac \, dx + C$
: $\int\frac \, dx= -\frac-\frac\int\frac \, dx + C$
: $\int\frac \, dx= \frac\ln\left, \frac\-\frac\int\frac \, dx + C$

Integrands of the form ''x''^''m'' (''a'' + ''b x''^''n'')^''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.

:$\int x^m \left(a+b\,x^n\right)^p dx = \frac\,+\, \frac\int x^m \left(a+b\,x^n\right)^dx$

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

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

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

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

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

Integrands of the form (''A'' + ''B x'') (''a'' + ''b x'')^''m'' (''c'' + ''d x'')^''n'' (''e'' + ''f x'')^''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'', ''n'' 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 $(a+b\,x)^m (c+d\,x)^n (e+f\,x)^p$ by setting ''B'' to 0.

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

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

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

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

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

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

Integrands of the form ''x''^''m'' (''A'' + ''B x''^''n'') (''a'' + ''b x''^''n'')^''p'' (''c'' + ''d x''^''n'')^''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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Integrands of the form (''d'' + ''e x'')^''m'' (''a'' + ''b x'' + ''c x''²)^''p'' when ''b''² − 4 ''a c'' = 0

* 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 $\left(a+b\,x+c\,x^2\right)^p$ when $b^2-4\,a\,c=0$ by setting ''m'' to 0.

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

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

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

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

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

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

:$\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx= \frac\,+\, \frac \int (d+e\,x)^\left(a+b\,x+c\,x^2\right)^pdx$

:$\int (d+e\,x)^m\left(a+b\,x+c\,x^2\right)^pdx= -\frac\,+\, \frac \int (d+e\,x)^ \left(a+b\,x+c\,x^2\right)^pdx$

Integrands of the form (''d'' + ''e x'')^''m'' (''A'' + ''B x'') (''a'' + ''b x'' + ''c x''²)^''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 $\left(a+b\,x+c\,x^2\right)^p$ and $(d+e\,x)^m \left(a+b\,x+c\,x^2\right)^p$ by setting ''m'' and/or ''B'' to 0.

:$\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx= \frac\,+\, \fracp\,\cdot$

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

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

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

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

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

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

:$\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx= \frac\,+$

::$\frac\,\cdot$

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

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

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

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

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

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

Integrands of the form ''x''^''m'' (''a'' + ''b x''^''n'' + ''c x''^2''n'')^''p'' when ''b''² − 4 ''a c'' = 0

* 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 $\left(a+b\,x^n+c\,x^\right)^p$ when $b^2-4\,a\,c=0$ by setting ''m'' to 0.

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

:$\int x^m \left(a+b\,x^n+c\,x^\right)^p dx= \frac\,+\, \frac\,+\, \frac \int x^ \left(a+b\,x^n+c\,x^\right)^dx$

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

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

:$\int x^m \left(a+b\,x^n+c\,x^\right)^p dx= \frac\,+\, \frac\,+\, \frac \int x^m \left(a+b\,x^n+c\,x^\right)^dx$

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

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

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

Integrands of the form ''x''^''m'' (''A'' + ''B x''^''n'') (''a'' + ''b x''^''n'' + ''c x''^2''n'')^''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 $\left(a+b\,x^n+c\,x^\right)^p$ and $x^m \left(a+b\,x^n+c\,x^\right)^p$ by setting ''m'' and/or ''B'' to 0.

:$\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^\right)^pdx= \frac\,+\, \frac\,\cdot$

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

:$\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^\right)^pdx= \frac\,+\, \frac\,\cdot$

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

:$\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^\right)^pdx= \frac\,+\, \frac\,\cdot$

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

:$\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^\right)^pdx= -\frac\,+\, \frac\,\cdot$

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

:$\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^\right)^pdx= \frac\,-\, \frac\,\cdot$

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

:$\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^\right)^pdx= \frac\,+\, \frac\,\cdot$

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

References

{{Lists of integrals

Rational functions