Question

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

Answer 1

$I=\int \frac{xdx}{(x-1)(x^2+1)}dx$
Let $\frac{x}{(x-1)(x^2+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+1)}$
$x=A{x}^2+A+B{x}^2+Cx-Bx-C$
On solving, we get
$A=\frac{1}{2},B=-\frac{1}{2},C=\frac{1}{2}$
$I=\int \frac{\frac{1}{2}}{(x-1)}dx+\int \frac{-\frac{1}{2}x+\frac{1}{2}}{(x^2+1)}dx$
$I_1=\frac{1}{2}\log |x-1|+c_1$
$I_2=\frac{1}{2}\int \frac{-x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx$
$=-\frac{1}{4}\int \frac{2x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx$
$=-\frac{1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}|x|+c_2$
$I=I_1+I_2$
$=\frac{1}{2}\log |x-1|-\frac{1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}|x|+c$
Where, $c=(c_1+c_2)$ is constant of Integration.