Question

Evaluate the following integral:

$\int \frac{x+3}{(x-1)(x^2+1)}dx$

Answer 1

Consider, $I=\int \frac{x+3}{(x-1)(x^2+1)}dx$

$\frac{x+3}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$

$\Rightarrow x+3=A(x^2+1)+(Bx+C)(x-1)$

$\Rightarrow x+3=(A+B)x^2+(C-B)x+(A-C)$

$\Rightarrow A+B=0$

and $C-B=1$

and $A-C=3$

Solving these equations we get-

$A=2,B=-2,C=-1$

$\int \frac{x+3}{(x-1)(x^2+1)}dx$

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

$=2\ln |x-1|-\int \frac{2x}{x^2+1}dx-\int \frac{1}{x^2+1}dx$

Let $z=x^2+1+\Rightarrow dx=2xdx$

$\Rightarrow \int \frac{2xdx}{x^2+1}=\frac{dz}{z}=\ln z=\ln (x^2+1)$

Hence integration becomes-

$I=2\ln |x-1|-\ln (x^2+1)-{\tan }^{-1}x+c$