<s>[INST] 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?
 [/INST] 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.
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