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?
 A: $$\frac{x}{\sqrt{x^2-1}} - \frac{x}{(x^2-1)^{3/2} \left(1+\frac{1}{x^2-1}\right)} = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1} (x^2-1+1)}$$
$$ = \frac{x}{\sqrt{x^2-1}} - \frac{x}{x^2 \sqrt{x^2-1}} = \frac{x^3 - x}{x^2 \sqrt{x^2-1}} = \frac{x (x^2-1)}{x^2 \sqrt{x^2-1}} = \frac{\sqrt{x^2-1}}{x}$$
A: You can use u-substitution
$u = \sqrt{x^2-1}$ then we have $du = \frac{x}{\sqrt{x^2-1}} dx$
then you will have $ \frac{u}{x}du = dx$
your integral now will be $\int \frac{u^2}{u^2+1} du$ which is much easier to solve.
A: arcsec$(x)$ is the angle whose secant is $x$.  I'm guessing you may be more familiar with the derivative of inverse cosine.  The angle that has a secant of $x$ has a cosine of $\frac1x$.
Also, your answer is more or less the same, just in a different form.  Draw a right triangle and call one of the acute angles $A$.  Secant is $\frac{\text{hypoteneuse}}{\text{adjacent}}$.  If the length of the hypoteneuse is $x$ and the length of the side adjacent to $A$ is $1$, then $\sec(A)=x$.  By the Pythagorean Theorem, the other leg has length $\sqrt{x^2-1}$.  From this triangle, we get $\tan(A)=\tan(\sec^{-1}x)=\sqrt{x^2-1}$.
A: Calculating this integral can be done in this way:
$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$
A: \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}
