How does this
$\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}}$
become
$\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}}$
According to my notes I let
u = x^2 + 4
x^2 = u - 4
but I do not see how that gives me the above.
$\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}} \, dx$
$\displaystyle u = x^2 + 4$
$\displaystyle du = 2x \, dx$
$\displaystyle x^2 = u - 4$
$\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}} \cdot 2x \, dx$
substitute ...
$\displaystyle \frac{1}{2}\int\frac{(u-4)^2}{\sqrt[3]{u}} \, du$