Question

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

Answer 1

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

$⟹A(x^2+1)+(Bx+C)(x-1)=x+3$

on solving this we get $A=2,B=-2,C=-1$

$\int \frac{x+3}{(x-3)(x^2+1)}dx=2\int \frac{dx}{x-1}-\int \frac{2xdx}{x^2+1}-\int \frac{dx}{x^2+1}=2\ln (x-1)-\ln (x^2+1)-{\text{tan}}^{-1}x+C$