Question

Evaluate : $\int \frac{dx}{x(x^2+1)}$

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

Answer 1

The correct option is A $-\frac{1}{2}\ln \left(\frac{x^2+1}{x^2}\right)+c$

Given, $\int \frac{dx}{x(x^2+1)}$
Resolving $\frac{1}{x(x^2+1)}$ into partial fractions
$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{(x^2+1)}$ ....(1)
$1=(A+B)x^2+$
$1=(A+B)x^2+Cx+A$
On comparing, we get
$A=1,C=0,B=-1$
Put these values in (1), we get
$\frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{(x^2+1)}$
Integrating both sides, we get
$\int \frac{dx}{x(x^2+1)}=\int \frac{dx}{x}-\int \frac{xdx}{(x^2+1)}$
$\Rightarrow \int \frac{dx}{x(x^2+1)}=\log x-\frac{1}{2}\log |1+x^2|+c$
$\int \frac{dx}{x(x^2+1)}=-\frac{1}{2}\log \left(\frac{x^2+1}{x^2}\right)+c$