# Evaluating $\int_0^{\sqrt{3}}{\frac{\sqrt{1+x^2}}{x}}\,dx$

Could someone please show me how to evaluate this integral (maybe doing all the steps)? $$\int_0^{\sqrt{3}}{\frac{\sqrt{1+x^2}}{x}}\,dx$$ I prefer if you avoid to follow the same method used by WolframAlpha (with $\csc$, $\sec$ ecc).
This is what I tried 'till now:

1. Substitution with $\sqrt{1+x^2} = u$ I obtained: $$\int{\frac{u}{\sqrt{u^2-1}}\frac{u}{\sqrt{u^2-1}}}\,du = \int{\frac{u^2}{u^2-1}}\,du$$ But not knowing how to continue, I tried another substitution with $u^2 - 1 = s$ and I obtained: $$\int{\frac{s+1}{s} \frac{1}{2\sqrt{s+1}} }\,ds = \frac{1}{2}\int{\frac{s+1}{s\sqrt{s+1}}}\,ds$$ But, again, not knowing how to continue I decided to ask here.

Thanks in advance for the help!

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Try wolframalpha.com for evaluating integrals. It even shows a step-by-step-solution! – FUZxxl Jun 21 '12 at 17:12
i guess there is a typo in the integral somewhere. $$\int_0^{\sqrt{3}}\frac{\sqrt{1+x^2}}{x}\,dx>\int_0^{\sqrt{3}}\frac{1}{x}\,dx =+\infty.$$ – vesszabo Jun 21 '12 at 20:33
@vesszabo: i was typing the same thing, but 2 minutes after you. I just noticed that. :-) I'll delete my post. – user 1618033 Jun 21 '12 at 20:47
It's good to notice that depending on integration interval one may use a nice trick that involves the inverse of the integrand that could be direclty integrated. – user 1618033 Jun 21 '12 at 21:10
@Chris Could you show me how? It seems interesting :) – Overflowh Jun 22 '12 at 11:11

For every $u \in \mathbb{R}\setminus\{-1,1\}$ we have $$\frac{u^2}{u^2-1}=\frac{u^2-1+1}{u^2-1}=1+\frac{1}{u^2-1} =1+\frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right),$$ and so $$\int\frac{u^2}{u^2-1}du=\int\left[1+\frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\right]du=u+\frac{1}{2}\ln\left|\frac{u-1}{u+1}\right| +C.$$

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Oh.. now it's clear! I could not understand how to obtain the partial fractions.. Thank you :) – Overflowh Jun 21 '12 at 17:15
Ehm.. I'm sorry but.. could you tell me how you obtained $1+\frac{1}{2}(\frac{1}{u-1}-\frac{1}{u+1})$? I'm trying, but I can't find a simple ad immediate way.. – Overflowh Jun 21 '12 at 17:21
This is known as the method of partial fractions. Any calculus textbook should have an explanation (at least for cases like this with linear factors). – GEdgar Jun 21 '12 at 18:13
$$\frac{1}{u^2-1}=\frac{1}{2}\frac{u+1-(u-1)}{u^2-1}=...$$ – Mercy King Jun 21 '12 at 18:32

Your first substitution was good, but the next one kind of got away from the solution. After you reach $\displaystyle\int \frac{u^2}{u^2-1} du$ apply partial fractions.

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$$\frac{u^2}{u^2-1}=1+\frac12\frac1{u-1}-\frac12\frac1{u+1}$$

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WA said the da same thing.. but I haven't immediately understand how to get this result.. – Overflowh Jun 21 '12 at 17:10
First get rid of the $1$, then decompose $1/(u^2-1)=1/((u-1)(u+1))$ into a linear combination of the simple fractions $1/(u\pm1)$. – Did Jun 21 '12 at 17:14
$\frac{u^2}{u^2-1} = \frac{u^2-1+1}{u^2-1} = 1 + \frac{1}{u^2-1}$. Then do partial fraction. This is the standard trick, called adding $0$. – N.U. Jun 21 '12 at 17:14
I'm sorry... could you tell me if there is a formula to do the partial fraction? – Overflowh Jun 21 '12 at 17:31
There is. See here. – Did Jun 21 '12 at 17:33