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
    Try http://wolframalpha.com for evaluating integrals. It even shows a step-by-step-solution!2012-06-21
  • 1
    i guess there is a typo in the integral somewhere. $$\int_0^{\sqrt{3}}\frac{\sqrt{1+x^2}}{x}\,dx>\int_0^{\sqrt{3}}\frac{1}{x}\,dx =+\infty.$$2012-06-21
  • 1
    @vesszabo: i was typing the same thing, but 2 minutes after you. I just noticed that. :-) I'll delete my post.2012-06-21
  • 1
    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.2012-06-21
  • 0
    @Chris Could you show me how? It seems interesting :)2012-06-22
  • 0
    @unNaturhal: look at vesszabo's comment. I think that integration interval is wrong, or?2012-06-22
  • 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

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

  • 0
    WA said the da same thing.. but I haven't immediately understand how to get this result..2012-06-21
  • 1
    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 $1/(u\pm1)$.2012-06-21
  • 1
    $\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 $0$.2012-06-21
  • 0
    I'm sorry... could you tell me if there is a formula to do the partial fraction?2012-06-21
  • 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
    Oh.. now it's clear! I could not understand how to obtain the partial fractions.. Thank you :)2012-06-21
  • 0
    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..2012-06-21
  • 1
    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).2012-06-21
  • 0
    $$\frac{1}{u^2-1}=\frac{1}{2}\frac{u+1-(u-1)}{u^2-1}=...$$2012-06-21