Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone show how to calculate $$\int\sqrt{1+u^2}\,du?$$ I can't calculate it.

share|improve this question
Calculate would be a better word. Sorry for confusing you. –  CALC-FATH Dec 3 '13 at 2:57
Stewart tells us to do a trig sub but $\sec^3\theta$ leaves a bit of work...Kaster's approach is much more efficient for whoever actually tries to carry out the computation til the end. –  1015 Dec 3 '13 at 3:10
I don't normally think of this integral as anything special, but having glanced at the responses below I'm filing this one away as a future teaching tool, since it so wonderfully demonstrates the versatility of integration tactics. –  David H Dec 3 '13 at 3:45

5 Answers 5

Setting $$ u=\sinh x:=\frac{e^x-e^{-x}}{2}, $$ we have \begin{eqnarray} \int\sqrt{1+u^2}\,du&=&\int\cosh x\sqrt{1+\sinh^2x}\,dx=\int\cosh^2x\,dx=\frac12\int(1+\cosh2x)\,dx\\ &=&\frac{x}{2}+\frac14\sinh2x+C=\frac{x}{2}+\frac12\sinh x\cosh x+C\\ &=&\frac12\ln(u+\sqrt{1+u^2})+\frac12u\sqrt{1+u^2}+C. \end{eqnarray}

share|improve this answer
While the tan substitution is the most obious one, I really like this one too. Because it is more (or should I say different) work. Versatility counts too. So upvote! –  imranfat Dec 3 '13 at 3:10

Let $u = \sinh x$, then \begin{align} \int \sqrt{1+u^2}du &= \int \cosh x \cdot \cosh x dx = \int \cosh^2x dx = \int \frac {1+\cosh 2x}2 dx = \\ &= \frac x2 + \frac {\sinh 2x}4 + C = \frac {\text{arcsinh } u}2 + \frac {u \sqrt{1+u^2}}2 + C \end{align}


To verify: WA

share|improve this answer

To integrate this we would use Trigonometric substitution. Recall your substitution rules for this. In this case, we would let $u = 1\tan(\theta)$ and then continue to find $d\theta$ so we may do the full substitution.

share|improve this answer
It'd be worth noting how you can figure out these substitution relationships by using the $sin^2 + cos^2 = 1$ identity. Then you don't have to memorize anything ;) –  GraphicsMuncher Dec 3 '13 at 3:01
I really would like to see how you'll proceed after that! –  Mercy Dec 3 '13 at 3:16
@Mercy It's kinda like your hyperbolic substitution, but ends up with the integral of $\sec^3x$. If you've either a) memorized the integral of $\sec^3x$, or b) become used to computing integrals like that, so it doesn't phase you, then you're all set. :) –  apnorton Dec 3 '13 at 3:33

Let $\displaystyle I = \int \sqrt{1+u^2}du = \int \sqrt{1+u^2}\cdot 1\;du$

Using Integration by parts

$\displaystyle I = \sqrt{1+u^2}\cdot u-\int \frac{u}{\sqrt{1+u^2}}\cdot udu = \sqrt{1+u^2}-\int\frac{(1+u^2)-1}{\sqrt{1+u^2}}du$

$\displaystyle I = \sqrt{1+u^2}\cdot u-I+\int\frac{1}{\sqrt{1+u^2}}du$

$\displaystyle 2I = \sqrt{1+u^2}\cdot u+J$

where $\displaystyle J = \int \frac{1}{\sqrt{1+u^2}}du$

for Calculation of $J$

Let $1+u^2 = v^2$ and $\displaystyle udu = vdv\Rightarrow \frac{du}{v} = \frac{dv}{u} = \frac{d(u+v)}{(u+v)}$

(above Using ratio and Proportion )

So $\displaystyle J = \int\frac{du}{v} = \int\frac{d(u+v)}{(u+v)} = \ln \left|u+v\right|+C$

So $\displaystyle J = \ln \left|u+\sqrt{1+u^2}\right|+C$

So $\displaystyle I = \frac{u}{2}\sqrt{1+u^2}+\frac{1}{2}\ln \left|u+\sqrt{1+u^2}\right|+D$

share|improve this answer

Let $ u = \tan \theta $. Then, $ \mathrm{d}u = \sec^2 \theta \, \mathrm{d}\theta $. Also, note that $$ \sqrt {1 + u^2} = \sqrt {1 + \tan^2 \theta} = \sec \theta. $$Use this to finish.

share|improve this answer
I really would like to see how you'll proceed after that! –  Mercy Dec 3 '13 at 3:16
@Mercy I simply did not show my work because the OP hasn't shown any him/herself. But it becomes simple Integration by Parts after that. –  Ahaan S. Rungta Dec 3 '13 at 3:18
Are you sure about that? –  Mercy Dec 3 '13 at 3:20
@Mercy Yes. I have tried it. –  Ahaan S. Rungta Dec 3 '13 at 3:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.