Answer
$\frac{x^{2}}{2}-\frac{1}{2}\ln(x^{2}+1)+tan^{-1}x+C$
Work Step by Step
long division gives
$\frac{x^{3}+1}{x^{2}+1}$ = $x-\frac{x-1}{x^{2}+1}$
so
$\int(\frac{x^{3}+1}{x^{2}+1})dx$ = $\int(x-\frac{x}{x^{2}+1}+\frac{1}{x^{2}+1})dx$ = $\frac{x^{2}}{2}-\frac{1}{2}\ln(x^{2}+1)+tan^{-1}x+C$
