Question

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

Answer 1

$\int \frac{x^2+3x+2}{{(x^2+1)}^2(x+1)}dxLet,f\left(x\right)=\frac{x^2+3x+2}{{(x^2+1)}^2(x+1)}=\frac{(x+1)(x+2)}{{(x^2+1)}^2(x+1)}=\frac{x+2}{{(x^2+1)}^2}Hence,\frac{x+2}{{(x^2+1)}^2}=\frac{A}{(x^2+1)}+\frac{B}{{(x^2+1)}^2}\Rightarrow x+2=A(x^2+1)+B\Rightarrow x+2=A{x}^2+A+B$

Comparing the coafficient of x2 we get A=0

and also the contant term, we get A+B=2

B=2 (since A=0)

$f\left(x\right)=\frac{x+2}{{(x^2+1)}^2}=\frac{0}{(x^2+1)}+\frac{2}{{(x^2+1)}^2}=\frac{2}{{(x^2+1)}^2}Now,\int \frac{x+2}{{(x^2+1)}^2}=\int \frac{2}{{(x^2+1)}^2}=\frac{1}{x(x^2+1)}\therefore \int \frac{x^2+3x+2}{{(x^2+1)}^2(x+1)}dx=\frac{1}{x(x^2+1)}$