Answer
$\dfrac{1}{4}\ln(x^4+2x^2+1)+\dfrac{x}{2(x^2+1)}+\dfrac{\tan^{-1}x}{2}+C$
Work Step by Step
Let $I=\int\dfrac{x^3+x+1}{(x^2+1)^2}dx=\int\dfrac{x^3+x}{x^4+2x^2+1}dx+\int\dfrac{1}{(x^2+1)^2}dx$
Suppose $u=x^4+2x^2+1\implies du=(4x^3+4x)dx$
or, $I=\dfrac{1}{4}\int\dfrac{1}{u}du+[\dfrac{x}{2(x^2+1)}+\dfrac{1}{2}\tan^{-1}x]+C \\=\dfrac{1}{4}\ln|u|+\dfrac{x}{2(x^2+1)}+\dfrac{1}{2}\tan^{-1}x+C\\=\dfrac{1}{4}\ln(x^4+2x^2+1)+\dfrac{x}{2(x^2+1)}+\dfrac{\tan^{-1}x}{2}+C$
