Question

Evaluate:

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

• A. $\frac{1}{2}\log \left|\frac{x^2-x^3}{1+x}\right|+C$
• B. $\log \left|\frac{1-x}{x(1+x)}\right|+C$
• C. $\log \left|x-x^3\right|+C$
• D. $\frac{1}{2}\log \left|\frac{x^2}{1-x}\right|+C$

Answer 1

The correct option is A $\frac{1}{2}\log \left|\frac{x^2-x^3}{1+x}\right|+C$

$\int \frac{dx}{x-x^3}=\int \frac{dx}{x(1+x)(1-x)}$

Breaking it into partial fractions

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

$1=A(1+x)(1-x)+B\left(x\right)(1-x)+Cx(1+x)$

Put $x=0$

$A=1$

Put $x=1$

$C=\frac{1}{2}$

Put $x=-1$

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

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

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

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

$=\frac{1}{2}log\left|\frac{x^2-x^3}{1+x}\right|+c$

A is correct.