Question

Choose the correct answer in each of the question.
$\int \frac{dx}{x(x^2+1)}$ equals
$\left(a\right)log|x|-\frac{1}{2}log(x^2+1)+C\left(b\right)log|x|+\frac{1}{2}log(x^2+1)+C\left(c\right)-log|x|+\frac{1}{2}log(x^2+1)+C\left(d\right)\frac{1}{2}log|x|+log(x^2+1)+C$

Answer 1

Let $\frac{1}{x(x^2+1)}=\frac{A}{2}+\frac{Bx+C}{x^2+1}\Rightarrow 1=A(x^2+1)+(Bx+C)x$
On equating the coefficients of $x^2$, x and constant term on both sides, we get
A+B=0, C =0 and A =1
On solving these equations, we get
A=1, B=-1 and C=0
$\therefore \frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{x^2+1}\therefore \int \frac{1}{x(x^2+1)}dx=\int \{\frac{1}{x}-\frac{x}{x^2+1}dx\}=log|x|-\frac{1}{2}log(x^2+1)+C[\text{Let}{x}^2+1=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}\therefore \int \frac{x}{x^2+1}dx=\int \frac{1}{t}\frac{dt}{2}=\frac{1}{2}logt]$
So, the option (a) is correct.