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?

Calculating this integral can be done in this way:
$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$
