antiderivative of rational function

The most notable real functions, which can be integrated in a closed form, are the rational functions:

Theorem.  The antiderivative of a rational function is always expressible in a closed form, which only can comprise, except a rational expression summand, summands of logarithms and arcustangents of rational functions.

One can justify the theorem by using the general form of the (unique) partial fraction decomposition

$R(x)=\mathit{\hspace{1em}}H(x)+$ ${\displaystyle \sum _{i=1}^m}\left({\displaystyle \frac{A_{i1}}{x-a_i}}+{\displaystyle \frac{A_{i2}}{{(x-a_i)}^2}}+\mathrm{\dots }+{\displaystyle \frac{A_{i{\mu }_i}}{{(x-a_i)}^{{\mu }_i}}}\right)$

$+$ ${\displaystyle \sum _{j=1}^n}\left({\displaystyle \frac{B_{j1}x+C_{j1}}{x^2+2p_{j}x+q_j}}+{\displaystyle \frac{B_{j2}x+C_{j2}}{{(x^2+2p_{j}x+q_j)}^2}}+\mathrm{\dots }+{\displaystyle \frac{B_{j{\nu }_j}x+C_{j{\nu }_j}}{{(x^2+2p_{j}x+q_j)}^{{\nu }_j}}}\right),$

of the rational function $R(x)$ ; here, $H(x)$ is a polynomial, the first sum expression is determined by the real zeroes $a_i$ of the denominator of $R(x)$, the second sum is determined by the real quadratic prime factors $x^2+2p_{j}x+q_j$ of the denominator (which have no real zeroes).

The addends of the form ${\displaystyle \frac{A}{{(x-a)}^r}}$ in the first sum are integrated directly, giving

${\displaystyle \int }{\displaystyle \frac{A}{x-a}}dx=A\ln |x-a|+\text{constant}\mathit{\hspace{1em}\hspace{1em}}(r=1)$ (1)

and

${\displaystyle \int }{\displaystyle \frac{A}{{(x-a)}^r}}dx=-{\displaystyle \frac{A}{r-1}}\cdot {\displaystyle \frac{1}{{(x-a)}^{r-1}}}+\text{constant}\mathit{\hspace{1em}\hspace{1em}}(r>1).$ (2)

The remaining partial fractions are of the form ${\displaystyle \frac{Bx+C}{{(x^2+2px+q)}^s}}$ where    and $s$ is a positive integer.  Now we may write

$$x^2+2px+q={(x+p)}^2+q-p^2=(q-p^2)\left[1+{\left(\frac{x+p}{\sqrt{q-p^2}}\right)}^2\right]$$

and make the substitution

${\displaystyle \frac{x+p}{\sqrt{q-p^2}}}=t,$ (3)

i.e.  $x=t\sqrt{q-p^2}-p$,  getting

${\displaystyle \int \frac{Bx+C}{{(x^2+2px+q)}^s}𝑑x}={\displaystyle \int \frac{Et+F}{{(1+t^2)}^s}𝑑t}=E{\displaystyle \int \frac{tdt}{{(1+t^2)}^s}}+F{\displaystyle \int \frac{dt}{{(1+t^2)}^s}}$ (4)

where $E$ and $F$ are certain constants.  In the case  $s=1$  we have

${\displaystyle \int \frac{tdt}{1+t^2}}={\displaystyle \frac{1}{2}}\ln (1+t^2)+\text{constant}$ (5)

and in the case  $s>1$

${\displaystyle \int \frac{tdt}{{(1+t^2)}^s}}=-{\displaystyle \frac{1}{2(s-1)}}\cdot {\displaystyle \frac{1}{{(1+t^2)}^{s-1}}}+\text{constant}.$ (6)

The latter addend of the right hand side of (4) is for  $s=1$  got from

${\displaystyle \int \frac{dt}{1+t^2}}=\arctan t+\text{constant}$ (7)

and for the cases $s>1$ on may first write

$$\int \frac{dt}{{(1+t^2)}^s}=\int \frac{(1+t^2)-t^2}{{(1+t^2)}^s}𝑑t=\int \frac{dt}{{(1+t^2)}^{s-1}}-\int t\cdot \frac{tdt}{{(1+t^2)}^s}.$$

Using integration by parts in the last integral, this equation can be converted into the reduction formula

${\displaystyle \int \frac{dt}{{(1+t^2)}^s}}={\displaystyle \frac{1}{2s-2}}\cdot {\displaystyle \frac{t}{{(1+t^2)}^{s-1}}}+{\displaystyle \frac{2n-3}{2n-2}}{\displaystyle \int \frac{dt}{{(1+t^2)}^{s-1}}}.$ (8)

The assertion of the theorem follows from (1), …, (8).

Example.

$$\int \frac{dx}{1+x^4}=\frac{1}{4\sqrt{2}}\ln \frac{1+x\sqrt{2}+x^2}{1-x\sqrt{2}+x^2}+\frac{1}{2\sqrt{2}}\arctan \frac{x\sqrt{2}}{1-x^2}+C$$

Title	antiderivative of rational function
Canonical name	AntiderivativeOfRationalFunction
Date of creation	2013-03-22 19:21:38
Last modified on	2013-03-22 19:21:38
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	19
Author	pahio (2872)
Entry type	Theorem
Classification	msc 26A36
Synonym	integration of rational functions
Related topic	IntegrationTechniques