Question

Evaluate:

$\int \frac{1}{x-x^3}dx$

• A. $\frac{1}{2}\log \left|\frac{1-x^2}{x^2}\right|+c$
• B. $\log \left|\frac{1-x}{x(1+x)}\right|+c$
• C. $\log \left|x(1-x^2)\right|+c$
• D. $\frac{1}{2}\log \left|\frac{x^2}{x^2-1}\right|+c$

Answer 1

The correct option is D $\frac{1}{2}\log \left|\frac{x^2}{x^2-1}\right|+c$

$\int \frac{dx}{x-x^3}$

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

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

Solving by the method of partial fractions,

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

$\Rightarrow 1=A(1-x)(1+x)+Bx(1+x)+Cx(1-x)$

Put $x=0\Rightarrow A=1$

Put $x=1\Rightarrow 2B=1$ or $B=\frac{1}{2}$

Put $x=-1\Rightarrow -2C=1$ or $C=-\frac{1}{2}$

Now,$\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2}\frac{1}{1-x}-\frac{1}{2}\frac{1}{1+x}$

Hence $\int \frac{dx}{x-x^3}$

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

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

$=\frac{2}{2}\log \left|x\right|-\frac{1}{2}\log \left|(1-x^2)\right|+c$

$=\frac{1}{2}\log \left|x^2\right|-\frac{1}{2}\log \left|(1-x^2)\right|+c$

$=\frac{1}{2}\log \left|\frac{x^2}{1-x^2}\right|+c$

$=\frac{1}{2}\log \left|\frac{x^2}{x^2-1}\right|+c$ for $x\ne 0,1$