Answer
$4 \ln |x|-\dfrac{1}{2} \ln (x^2+1)+4 \tan^{-1} x+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Solve $\int \dfrac{3x^2+4x+4}{x^3+x}dx$
Write the integral into partial fractions as follows: $\int \dfrac{3x^2+4x+4}{x^3+x}dx=\int \dfrac{4}{x} dx+\int \dfrac{-(x+4)}{(x^2+1)}dx$
Plug $x^2+1=a \implies da=(2x) dx$
This implies that $4 \ln |x|-\dfrac{1}{2}\int \dfrac{da}{a}+(4) \int \dfrac{dx}{(x^2+1)}=4 \ln |x|-\dfrac{1}{2} \ln (x^2+1)+4 \tan^{-1} x+c$
