<s>[INST] How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$? I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$
I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.
 [/INST] Setting $u=\sqrt{x-1}$ gives you
$$ \int \frac{2u \, du}{(1+u^2)^2u}= 2\int \frac{du}{(1+u^2)^2} \tag{1}$$
There are two ways to do this: the first is to change variables again to $u=\tan{\theta}$, so $du=\sec^2{\theta} \, d\theta$:
$$ \begin{align*}
2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} &= 2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} \\
&= \int 2\cos^2{\theta} \, d\theta \\
&= \int (1+\cos{2\theta}) d\theta, \end{align*}$$
using the double-angle formula $\cos{2\theta}=2\cos^2{\theta}-1$.
$$ \int (1+\cos{2\theta}) = \theta + \frac{1}{2}\sin{2\theta}, $$
and substituting back, $\theta = \arctan{\sqrt{x-1}}$, and in particular,
$$ \int \frac{dx}{x^2\sqrt{x-1}} = \frac{1}{2} \sin{(2\arctan{\sqrt{x-1}})} + \arctan{\sqrt{x-1}}. $$
The first term can be simplified using the formula $\sin{(2 \arctan{t})} = \frac{2t}{1+t^2}$ to give the answer.
Alternatively:
Having got to (1), we can write $1=(1+u^2)-u^2$ integrate by parts:
$$ 2\int \frac{(1+u^2)-u^2}{(1+u^2)^2}\,du = \int \frac{2du}{1+u^2} - \int u\frac{2u}{(1+u^2)^2} \, du \\
= \int \frac{2du}{1+u^2} + \frac{u}{1+u^2} - \int \frac{du}{1+u^2} \, du, $$
and then proceed to substitute $u=\tan{\theta}$ in the first one to get the $\arctan$ term. This way, we avoided having to use lots of double angle formulae.
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