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# Integrate the rational functions$\frac{1}{x^4-1}$

$\begin{array}{1 1} \frac{1}{4} \log \big | \frac {x-1}{x+1} \big | -\frac{1}{2}\tan^{-1}x+c \\ \frac{1}{4} \log \big | \frac {x+1}{x-1} \big | -\frac{1}{4}\tan^{-1}x+c \\\frac{1}{4} \log \big | \frac {x^2}{x-1} \big | -\frac{1}{2}\tan^{-1}x+c \\ \frac{1}{4} \log \big | \frac {x^2}{x+1} \big | +\frac{1}{2}\tan^{-1}x+c\end{array}$

Can you answer this question?

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x+a)(x^2+b)}$
• $\;$Form of the partial function$\frac{A}{(x+a)}+\frac{Bx+C}{(x^2+b)}$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
• $(iii)\;\int\frac{dx}{(x^2+b}=\frac{1}{b}\tan^{-1}\big(\frac{x}{b}\big)+c.$
Given:$I=\int\frac{1}{x^4-1}dx.$

$\frac{1}{x^4-1}=\frac{1}{(x^2-1)(x^2+1)}=\frac{1}{(x+1)(x-1)(x^2+1)}.$

Therefore $\frac{1}{(x+1)(x-1)(x^2+1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{(x+1)}{(x^2+1)}.$

$\Rightarrow 1=A(x-1)(x^2+1)+B(x+1)(x^2+1)+(x+D)(x+1)(x-1).$

$\;\;\;=A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+(x+D)(x^2-1).$

Equating the coefficients of $x^3$,

0=A+B+C------(1)

Equating the coefficients of $x^2$,

0=-A+B+D-----(2)

Equating the coefficients of x,

0=A+B-C-------(3)

Equating the constant terms,

1=-A+B-D-----(4)

Let us take equ(1) and equ(2) and eliminate A by adding them.
A+B+C=0
-A+B+D=0
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2B+C+D=0------(5)
Let us take equ(2) and equ(3) and eliminate A by adding then,
-A+B+D=0
A+B-C=0
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2B+D-C=0------(6)
Now add (5) and (6) to eliminate C
2B+C+D=0
2B-C+D=0
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4B+2D=0
$\Rightarrow$ 2B+D=0------(7)

Now subtract equ(1) and equ(4) to elliminate A and B

-A+B+D=0

-A+B-D=1

2D=-1$\Rightarrow D=\frac{-1}{2}.$

Substitute for D in equ(7)

2B+D=0

2B-$\frac{1}{2}$=0.

2B=$\frac{1}{2}\Rightarrow B=1/4.$

Substitute for B and D in equ(6)

2(1/4)+(-1/2)-C=0$\Rightarrow C=0.$

Now substitute for B and C in equ(1)

A+B+C=0.

A+(1/4)+0=0.

A=-1/4.

Hence A=-1/4,B=1/4,C=0 and D=-1/2.

Now substitute for A,B,C and D in I

$\frac{1}{(x-1)(x+1)(x^2+1)}=-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+\frac{0-1/2}{(x^2+1)}$.

$I=\frac{-1}{4}\int\frac{dx}{(x+1)}+\frac{1}{4}\int\frac{dx}{(x-1)}-\frac{-1}{2}\int\frac{dx}{(x^2+1)}.$

On integrating we get,

$\;\;\;=\frac{-1}{4}log|x+1|+\frac{1}{4}log|x-1|-\frac{1}{2}\tan^{-1}x+c.$

log a-log b=log$\mid\frac{a}{b}\mid$,similarly

$\;\;\;=\frac{1}{4}log\frac{|x-1|}{|x+1|}-\frac{1}{2}\tan^{-1}x+C.$

answered Feb 6, 2013
edited Jul 21, 2013