$\displaystyle \int{x(2-x^2)^3}\,dx$
How could I get the -1/2???
I think what you want is
$\displaystyle
\int x(2-x^2)^3\,dx = -\frac{1}{2}\int (2-x^2)^3\, d(2-x^2)
$
but it's really better if you use a substitution. Let $\displaystyle u = 2-x^2$ (the hard part of this). Then $\displaystyle du = -2 x \,dx$ so $\displaystyle x\,dx = -\frac{1}{2}\,du$ so the original problem becomes
$\displaystyle
-\frac{1}{2} \int u^3\,du
$