Question

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

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

Answer 1

The correct option is A $\log |x|-\frac{1}{2}\log (x^2+1)+C$

Let $\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
$\Rightarrow 1=A(x^2+1)+(Bx+C)x$
Equating the coefficients of $x^2,x$, the constant term, we obtain
$A+B=0,C=0,A=1$
On solving these equations, we obtain
$A=1,B=-1$, and $C=0$
$\therefore \frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{x^2+1}$
$\Rightarrow \int \frac{1}{x(x^2+1)}dx=\int \{\frac{1}{x}-\frac{x}{x^2+1}\}dx$
$=\log |x|-\frac{1}{2}\log |x^2+1|+C$