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?

Solution 1:

$$\frac{x}{\sqrt{x^2-1}} - \frac{x}{(x^2-1)^{3/2} \left(1+\frac{1}{x^2-1}\right)} = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1} (x^2-1+1)}$$ $$ = \frac{x}{\sqrt{x^2-1}} - \frac{x}{x^2 \sqrt{x^2-1}} = \frac{x^3 - x}{x^2 \sqrt{x^2-1}} = \frac{x (x^2-1)}{x^2 \sqrt{x^2-1}} = \frac{\sqrt{x^2-1}}{x}$$

Solution 2:

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 $\int \frac{u^2}{u^2+1} dx$ which is much easier to solve.

Solution 3:

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 $\tan(A)=\tan(\sec^{-1}x)=\sqrt{x^2-1}$.