Question

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

Answer 1

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

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

Put $x=\tan y⟹dx={\text{sec}}^{2}ydy$

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

$=\frac{1}{4}\ln (x^2+1)+\frac{1}{2}{\text{sec}}^{-1}x-\frac{1}{2}\ln (x+1)+y+\frac{\tan y}{1+{\tan }^{2}y}$

$=\frac{1}{4}\ln (x^2+1)+\frac{1}{2}{\text{sec}}^{-1}x-\frac{1}{2}\ln (x+1)+{\text{tan}}^{-1}x+\frac{x}{x^2+1}$