Answer
$\int_{-1}^{1} x(x^{2}+1)^3dx = 0$
Work Step by Step
Let $ u = x^2 + 1, du = 2x dx$
$\int_{-1}^{1} x(x^{2}+1)^3dx $
$= \frac{1}{2}\int_{-1}^1 {(x^{2}+1)}^{3}(2x)dx$
$= \frac{1}{2} \int_{-1}^1 u^{3}du$
$= \frac{1}{2} [\frac{u^4}{4}]_{-1}^1 $
$=\frac{1}{8} [(x^{2}+1)^4]_{-1}^1 $
$=\frac{1}{8}[(2^4)-(2^4)] $
$=\frac{1}{8}[0] $
$=0$
