Question

Integrate the rational function $\frac{x}{(x^2+1)(x-1)}$

Answer 1

Let $\frac{x}{(x^2+1)(x-1)}=\frac{Ax+B}{(x^2+1)}+\frac{C}{(x-1)}$

$\Rightarrow x=(Ax+B)(x-1)+C(x^2+1)$

$\Rightarrow x=A{x}^2-Ax+Bx-B+C{x}^2+C$

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

$A=-\frac{1}{2},B=\frac{1}{2}$, and $C=\frac{1}{2}$

From equation (1), we obtain

$\therefore \frac{x}{(x^2+1)(x-1)}=\frac{(-\frac{1}{2}x+\frac{1}{2})}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}$

$\Rightarrow \int \frac{x}{(x^2+1)(x-1)}=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x-1}dx$

$=-\frac{1}{4}\int \frac{2x}{x^2+1}dx+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log |x-1|+C$

Consider $\int \frac{2x}{x^2+1}dx$, let $(x^2+1)=t\Rightarrow 2xdx=dt$

$\Rightarrow \int \frac{2x}{x^2+1}dx=\int \frac{dt}{t}=\log |t|=\log |x^2+1|$

$\therefore \int \frac{x}{(x^2+1)(x-1)}=-\frac{1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log |x-1|+C$

$=\frac{1}{2}\log |x-1|-\frac{1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}x+C$