Solving $\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$ This is the last indefinite integral I am attempting before proceeding to definite integrals study. Any ideas how to solve it?
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$$
My attempt: (not the longest one... :P)
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx = \int \frac{\sqrt{(x-2)^3}}{x+1} dx = \left(\left(u=x-2,du=dx \right)\right) =\int\frac{\sqrt{u^3}}{u+3}du$$
But that substitution does not really make it simpler for me. If the expression inside the numerator's root was of the form $ax^2 + bx + c$ then I could apply one of Euler's substitution. But it's polynomial of third power.
How to deal with it?
 A: Consider 
$$I = \int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx = \int \frac{\sqrt{(x-2)^3}}{x+1} dx$$
Let $u = x -2 \implies du = dx $
$$I = \int\frac{u^\frac{3}{2}}{u + 3}du$$
Let $u = 3\tan^2t \implies du = 6\tan t \sec^2t dt$
$$I = 6\sqrt{27}\int\frac{\tan^4t\sec^2t}{3\sec^2t}dt$$
$$ = 6\sqrt{3}\int\tan^4t \ dt$$
$$ = 6\sqrt{3}\int\tan^2t(\sec^2t - 1)\ dt = 6\sqrt{3}\frac{\tan^3t}{3} - 6\sqrt{3}\int\tan^2 t\ dt$$
$$ = 2\sqrt{3}\tan^3t - 6\sqrt{3}\int(\sec^2t - 1)\ dt$$
$$ = 2\sqrt3\tan^3t - 6\sqrt{3}\tan t + 6\sqrt3t$$
$$I = 2\sqrt3(\tan^3t - 3\tan t + 3t)$$
Now $3\tan^2t = u = x-2 \implies \tan t = \left(\frac{x-2}{3}\right)^\frac{1}{2}$
$$I = 2\sqrt3 \left(\frac{x-2}{3}\right)^\frac{1}{2} \left(\frac{x-2}{3} - 3 \right) + 6\sqrt3\arctan\left(\sqrt{\frac{x-2}{3}}\right)$$
$$ I = \boxed{\frac{2}{3}(x-11)\sqrt{x-2} + 6\sqrt3\arctan\left(\sqrt{\frac{x-2}{3}}\right)}$$
Note: Wolfram Mathematica gives the same result as one of the alternate forms.
A: So you have brought the ques down to solving integral of u^(3/2)/(u+3)
Now simply put u= z^2 so after taking constants out it becomes
z^4/(z^2+3)= (z^4 - 9 + 9)/(z^2+3)
=(z^2 - 3)(z^2+3)/(z^2+3) + 9/(z^2+3)
= z^2 - 3 + 9/(z^2+3)
Here, z^2 - 3 is very easy to solve. For 9/(z^2+3) simply put z= sqrt(3).tan(u) and the integral becomes elementary.
