Trigonometric substitution integral Trying to work around this with trig substitution, but end up with messy powers on sines and cosines... It should be simple use of trigonometric properties, but I seem to be tripping somewhere.
$$\int x^5\sqrt{x^2+4}dx $$
Thanks.

It comes out pretty well if you put $x=2\tan\theta$. Doing it carefully, remove a $\tan\theta\sec\theta$ and everything else can be expressed as a polynomial in $\sec\theta$, hence it's easily done by substitution.
But a more "efficient" substitution is $u=\sqrt{x^2+4}$, or $u^2=x^2+4$. Then $2u\,du = 2x\,dx$ andthe integral becomes
$$\int (u^2-4)^2 u^2\,du\,.$$
