<s>[INST] How to find reduction formula for $I_n=\int\frac{1}{x^n \sqrt{x^2-1}}dx$ I have tried manipulating the integral but couldn't come up with an answer that matched textbook answer.
I just need a hint. Thanks in advance.
$\int\frac{dx}{x^n \sqrt{x^2-1}}$
This is what I tried:
Let $I_{n-2}=\int\frac{dx}{x^{n-2} \sqrt{x^2-1}}$
Then I integrated by parts.
The answer given in textbook is
$$(n-1)I=\frac{\sqrt{x^2-1}}{x^{n-1}} +(n-2)I_{n-2}$$
 [/INST] Reduce the integral as follows
\begin{align}
I_n= &\int\frac{1}{x^n \sqrt{x^2-1}}dx \\
=& \int\frac{(1-x^2)+x^2}{x^n \sqrt{x^2-1}}dx
=-\int \frac{\sqrt{x^2-1}}{x^n }dx + I_{n-2} \\
=&\ \frac1{n-1}\int \sqrt{x^2-1} \>d\left(\frac{1}{x^{n-1} }\right)+ I_{n-2} \\
\overset{ibp}=&\ \frac1{n-1}\frac{\sqrt{x^2-1}}{x^{n-1} }+\frac{n-2}{n-1}I_{n-2}
\end{align}
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