Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# Integrate the function$\int\frac{1}{x-x^3}$

Can you answer this question?

## 1 Answer

0 votes
Toolbox:
• (i)If the integral function is of the form $\int\frac{dx}{(x-a)(x-b)}$ then the function can be resolved into partial fractions of th form $\frac{A}{(x-a)}+\frac{B}{(x-b)}$ and then solved
• (ii)$\int \frac{1}{(x-a)}dx=log |x-a|+c$
Given $I=\int\frac{1}{x-x^3}$

This can be written as

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

But $(1-x^2)=(1-x)(1+x)$

$I=\int \frac{1}{x(1-x)1+x)}dx$

Resolving into a particular fraction

Now $\frac{1}{x(1-x)1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}$

$=>1=A(1-x)(1+x)+B(x(1+x))+C((x)(1-x))$

Equating the coefficient of $x^2$,

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

Equating the coefficient of x,

$0=B+C$-----(2)

Equating the constant terms,

$1=A$-----(3)

Substituting for A in equation (1)

$0=-1+B+C$

$=>B+C=1$-----(4)

Add equ (2) and equ(4)

$B+C=0$

$B-C=1$
__________
$\qquad 2B=1$

Hence $B=\frac{1}{2}$

Substitute the value for B in equ(2)

$\frac{1}{2}+c=0$

$=>c=-\frac{1}{2}$

Hence $A=1,B=\frac{1}{2}\;and\;c=-\frac{1}{2}$

Substituting for A,B and C in I we get

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

On seperating the terms we get,

$log x+(-\frac{1}{2})log|1-x|-\frac{1}{2}log|1+x|+c$

On taking $\frac{1}{2}$ as the common factor,

$=\frac{2log x-log|1-x|-log|1+x|}{2}$

We know $2logx=log x^2$ and

$log|1-x|+log|1+x|=log[(1-x)(1+x)]=log(1-x^2)$

Therefore $I=\frac{1}{2}log x^2-[\frac{1}{2}log(1+x) \times (1-x)]$

Therefore $\int \frac{dx}{x-x^3}=\frac{1}{2} log \bigg|\frac{x^2}{1-x^2}\bigg|+c$

answered Feb 15, 2013 by

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer