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Integrate the function\[\int\frac{1}{x-x^3}\]

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  • (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
But $(1-x^2)=(1-x)(1+x)$
$I=\int \frac{1}{x(1-x)1+x)}dx$
Resolving into a particular fraction
Now $\frac{1}{x(1-x)1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}$
Equating the coefficient of $x^2$,
Equating the coefficient of x,
Equating the constant terms,
Substituting for A in equation (1)
Add equ (2) and equ(4)
$\qquad 2B=1$
Hence $B=\frac{1}{2}$
Substitute the value for B in equ(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
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 meena.p