How do I calculate the integral when the denominator is a square root of a function?
e.g.
$\displaystyle \int \frac{u^3+1}{\sqrt{u^4+4u}} du$
This one you can solve using a substitution.
$\displaystyle \int{\frac{u^3 + 1}{\sqrt{u^4 + 4u}}\,du} = \frac{1}{4}\int{(u^4 + 4u)^{-\frac{1}{2}}(4u^3 + 4)\,du}$.
Now let $\displaystyle v = u^4 + 4u$ so that $\displaystyle dv = (4u^3 + 4)\,du$ and the integral becomes
$\displaystyle \frac{1}{4}\int{v^{-\frac{1}{2}}\,dv}$
$\displaystyle = \frac{1}{4}\left(\frac{v^{\frac{1}{2}}}{\frac{1}{2 }}\right) + C$
$\displaystyle = \frac{1}{4}\left(2\sqrt{v}\right) + C$
$\displaystyle = \frac{\sqrt{u^4 + 4u}}{2} + C$.
Please note that...
$\displaystyle \displaystyle \frac{u^{3}+1}{\sqrt{u^{4}+4\ u}} = \frac{1}{4}\ \frac{4\ u^{3}+4}{\sqrt{u^{4}+4\ u}}$ (1)
... and that...
$\displaystyle \displaystyle \frac{d}{d u}\ (u^{4}+4\ u) = 4\ u^{3} + 4$ (2)
At this point You are 'almost arrived' ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$