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.
 A: The change of variable
$$
\sqrt{x-1}=u\text{ or }x=u^2+1,\quad dx=2\,du
$$
transforms the integral into
$$
2\int\frac{du}{(u^2+1)^2}.
$$
Do you know how to do tjis one?
A: 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.
A: $$I=\int \dfrac{dx}{x^2 \sqrt{x-1}}$$
Let $x=\sec^2 t,dx=2\sec^2 t\tan t\, dt$:
$$ I=\int\dfrac{2\sec^2 t\tan t dt}{\sec^4t\tan t}=\int 2\cos^2tdt=\int(1+\cos 2t)dt=t+\dfrac12\sin2t+c=\arccos\dfrac1{\sqrt x}+\dfrac{\sqrt{x-1}}{x}+c$$
A: Substitute $u=\sqrt{x-1}$. Then $2du=\dfrac{1}{\sqrt{x-1}}dx$ and $x=u^2+1$. 
Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\int \dfrac{2}{(u^2+1)^2}du.$$
Put $U=\dfrac{1}{u^{2}+1}$ and $dV=du$. Then integration by parts,
$$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{u^{2}+1}+\int \dfrac{2u^2}{(u^2+1)^2}du\\=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{u^{2}+1}{(u^{2}+1)^2}-\dfrac{1}{(u^{2}+1)^2}\right)du\\
=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{1}{(u^{2}+1)}-\dfrac{1}{(u^{2}+1)^2}\right)du $$
Hence $$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\int \dfrac{1}{(u^{2}+1)}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\tan^{-1}(u).$$
Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\dfrac{\sqrt{x-1}}{x}+\tan^{-1}(\sqrt{x-1}).$$
A: \begin{align}
\int \dfrac{1}{x^2 \sqrt{x-1}}dx=&\int \frac{1}{\sqrt{x-1}}d\left(\frac{x-1}x\right)\\
= &\ \frac{\sqrt{x-1}}x +\int \frac{ d(\sqrt{x-1})}{x}
= \frac{\sqrt{x-1}}x +\tan^{-1} \sqrt{x-1}
\end{align}
