<s>[INST] Antiderivative fraction I'm would like to find the antiderivative of
$$\frac{x^2}{(x-1)^5}$$
I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving $$\frac{x²}{(x-1)^5}=\frac{1}{(x-1)^3}+\frac{2}{(x-1)^4}+\frac{1}{(x-1)^5}$$
Could you tell me how to proceed?
Thank you
 [/INST] Indeed you are correct that,
$$\frac{x^2}{(x-1)^5} = \frac{1}{(x-1)^3} + \frac{2}{(x-1)^4} + \frac{1}{(x-1)^5}$$
One can integrate each of these terms in turn.  I will do the first to help.  Let $u=x-1$ and $du=dx$ then,
$$ \int\frac{1}{(x-1)^3} dx = \int u^{-3} du  = -\frac{1}{2}u^{-2} = -\frac{1}{2}\frac{1}{(x-1)^2} + c$$
where $c$ is the constant of integration.  The final answer is,
$$ -\frac{1}{2}\frac{1}{(x-1)^2}  -\frac{2}{3}\frac{1}{(x-1)^3}  -\frac{1}{4}\frac{1}{(x-1)^4}+c $$
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