<s>[INST] Evaluate $\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x$ My try, using $x = \sec(u)$ substitution:
$$
\begin{eqnarray}
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\
&=& \int \tan^2(u) \mathrm{d}u \\
&=& \tan(u) - u + C \\
&=& \tan(arcsec(x)) - arcsec(x) + C
\end{eqnarray}
$$
However, according to Wolfram Alpha, the answer should be:
$$
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C
$$
When I derive this last answer I don't get back the integrand, but rather:
$$
\frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)}
$$
I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.
Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
 [/INST] \begin{aligned}
& \int \frac{\sqrt{x^{2}-1}}{x} d x \\
=& \int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
=& \int \frac{x^{2}-1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) \\
=& \sqrt{x^{2}-1}-\int{\frac{1}{x^{2}} d\left(\sqrt{x^{2}-1}\right)} \\
=& \sqrt{x^{2}-1}-\int \frac{d\left(\sqrt{x^{2}}-1\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1}\\
=& \sqrt{x^{2}-1}-\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
</s>