# 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?

• I know it is not your main question, but here is a good way to find the derivatives of the inverse trig functions if you ever don't know them. If $y=\sec^{-1} x$ then $x=\sec y$, and by implicit differentiation we fine $1=\sec y\tan y\frac{dy}{dx}\implies\frac{dy}{dx}=\frac{1}{\sec y\tan y}$. $\sec\sec^{-1} x=|x|$ and $\tan\sec^{-1} x=\sqrt{x^2-1}$. Therefore, the derivative of $\sec^{-1} x$ is $\frac{1}{|x|\sqrt{x^2-1}}$. Now you can check your answer. It is probably correct too. – solstafir Apr 16 '14 at 23:20
• – user85798 Apr 16 '14 at 23:22
• @solstafir Why is $\sec(arcsec(x)) = |x|$ with the absolute value? – dazedviper Apr 17 '14 at 12:36
• Sorry for the confusion. In general, $\sec\sec^{-1} x$ is not always positive. You have to use the absolute value here because of the $\tan y$. This is because $\sec y = \frac{1}{\cos y}$ and $\tan y = \frac{\sin y}{\cos y}$. For $y \in [0,\pi]$ (the range of $\sec^{-1} x$), $\sin y$ is always positive, but $\cos y$ is negative for $x \in (\frac{\pi}{2},\pi]$. But $\sec y$ and $\tan y$ are being multiplied, so the negatives cancel and the product is always positive. But in $\tan\sec^{-1} x=\sqrt{x^2-1}$ we always take the positive square root. $\sec\sec^{-1} x=|x|$ corrects for this. – solstafir Apr 17 '14 at 17:25
• – Martin Sleziak Aug 17 '18 at 13:24

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

• Thanks! I found this, a list of nice identities proved using similar constructions to yours. It may be useful to others. – dazedviper Apr 17 '14 at 12:50

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

Calculating this integral can be done in this way:

$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$

• Isn't $\int \frac{x^2-a^2}{x\sqrt{x^2-a^2}}\mathrm{d}x = \int \frac{x}{\sqrt{x^2-a^2}}\mathrm{d}x - \int \frac{a^2}{x\sqrt{x^2-a^2}}\mathrm{d}x$ ? Using $x = a\sec(u)$ substitution I get $\int \frac{x^2-a^2}{x}\mathrm{d}x = \sqrt{x^2-a^2} - aarcsec(\frac{x}{a}) + C$ – dazedviper Apr 17 '14 at 12:20
• OK! I corrected. Thank you! – medicu Apr 17 '14 at 13:06
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