Integration of $\int \sqrt\frac{x-1}{x^5}\ dx$? How would you integrate $\int \sqrt\frac{x-1}{x^5}\ dx$?
I've tried to split it up to integrate by parts, with $u = \sqrt{x-1}$ and $v' = \frac{1}{x^5}$. Then I get $-\frac{2}{3}x^{-3/2}(x-1)^{1/2} + \frac{1}{3} \int x^{-3/2}(x-1)^{-1/2}$. It seems to me that I need to further pursue integration by parts on the second integral, but that would seem to throw me into an endless loop. How can I proceed?
I am given that the answer to this is $\frac{2}{3}(1-\frac{1}{x})^{3/2} + C$. What might be the fastest way to achieve this answer?

Another approach:
Let $$I=\int \sqrt{\frac{x-1}{x^{5}}}dx$$so
\begin{eqnarray*}
I&=&\int \frac{\sqrt{x-1}}{x^{5/2}}dx\\
&=&2 \int \frac{\sqrt{u^{2}-1}}{u^4}du, \quad u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx\\
&=&2\int\sin^{2}(s)\cos(s)ds, \quad u=\sec(s), du=\tan(s)\sec(s)ds\\
&=&2\int p^{2}dp, \quad p=\sin(s), dp=\cos(s)ds\\
&=&\frac{2p^{3}}{3}+C\\
&=&\frac{2}{3}\left(\frac{x-1}{x^{5}} \right)^{3/2}x^{6}+C
\end{eqnarray*}
Therefore, $$\boxed{\int \sqrt{\frac{x-1}{x^{5}}}dx=\frac{2}{3}\left(\frac{x-1}{x^{5}} \right)^{3/2}x^{6}+C}$$
