Question

The value of the integral $\int \frac{x}{(x^2+1)(x-1)}dx$ is
(where $m$ is an arbitary constant)

• A. $\frac{1}{4}\ln |x+1|+\frac{1}{2}\ln |x^2+1|+\frac{1}{2}{\tan }^{-1}x+m$
• B. $\frac{1}{2}\ln |x-1|+\frac{1}{2}\ln |x+1|+\frac{1}{2}{\tan }^{-1}x+m$
• C. $\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+\frac{1}{2}{\tan }^{-1}x+m$
• D. $\frac{1}{2}\ln |x+1|+\frac{1}{4}\ln |x^2+1|+\frac{1}{2}{\tan }^{-1}x+m$

Answer 1

The correct option is C $\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+\frac{1}{2}{\tan }^{-1}x+m$

Let $\frac{x}{(x^2+1)(x-1)}=\frac{Ax+B}{(x^2+1)}+\frac{C}{(x-1)}\cdots \left(1\right)$
$x=(Ax+B)(x-1)+C(x^2+1)\Rightarrow x=A{x}^2-Ax+Bx-B+C{x}^2+C$
Equating the coefficients of $x^2,x$ and constant term, we obtain
$A+C=0$
$-A+B=1$
$-B+C=0$
On solving these equations, we obtain
$A=-\frac{1}{2},B=\frac{1}{2},C=\frac{1}{2}$
From equation $\left(1\right),$ we obtain
$\frac{x}{(x^2+1)(x-1)}=\frac{(-\frac{1}{2}x+\frac{1}{2})}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}$
Now,
$\Rightarrow \int \frac{x}{(x^2+1)(x-1)}dx=\frac{-1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}+\frac{1}{2}\int \frac{1}{x-1}dx=-\frac{1}{4}\int \frac{2x}{x^2+1}dx+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log |x-1|+C$

Considering $\int \frac{2x}{x^2+1}dx$
Let $(x^2+1)=t\Rightarrow 2xdx=dt$
$\Rightarrow \int \frac{2x}{x^2+1}dx=\int \frac{dt}{t}=\log |t|=\log |x^2+1|$

Therefore,
$\int \frac{x}{(x^2+1)(x-1)}dx=-\frac{1}{4}\ln |x^2+1|+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\ln |x-1|+m=\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+\frac{1}{2}{\tan }^{-1}x+m$
Where $m$ is an arbitary constant.