Answer
$-\dfrac{1}{3(x^3+3x+2)} $
Work Step by Step
We are given that $I=\int \dfrac{(x^2+1)}{(x^3+3x+2)^2} \ dx$
Let us consider that $u=x^3+3x+2 \implies dx=\dfrac{du}{3(x^2+1)}$
Now, we have $I=\int (x^2+1) u^{-2} [\dfrac{du}{3(x^2+1)}]$
or, $=\dfrac{1}{3} \int u^{-2} \ du$
or, $=-\dfrac{1}{3u} +C$
or, $=-\dfrac{1}{3(x^3+3x+2)} +C$
