3
$\begingroup$

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!

  • 0
    @Chris, No, it's right. It was an exam exercise that I'm doing to have practice..2012-06-22

3 Answers 3

1

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.

2

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

  • 1
    There is. See [here](http://en.wikipedia.org/wiki/Partial_fraction#Procedure).2012-06-21
2

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. $

  • 0
    $\frac{1}{u^2-1}=\frac{1}{2}\frac{u+1-(u-1)}{u^2-1}=...$2012-06-21