I want to find a quick way of evaluating $$\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$$

This problem appeared on the qualifying round of MIT's 2014 Integration Bee, which leads me to think there should be a shortish way (no more than three minutes by hand) to solve it. Examining the indefinite integral (thanks to WolframAlpha) hasn't particularly helped me:

$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=-\frac{2}{3} \left(\sqrt{x^2+1}-2x\right) \sqrt{x+\sqrt{x^2+1}}+C$$

The bounds on the integral hint at trigonometric substitution, but I got nowhere by trying $x=\tan u$.

I also noticed that we can transform the integral by multiplying $\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}$ in the first square root, but there didn't seem to be anything to do after that either.

  • $\begingroup$ I am unclear: is the question how to get WolframAlpha's expression or how to evaluate the definite integral? $\endgroup$ – robjohn Apr 19 '15 at 16:07
  • $\begingroup$ @robjohn It's how to get the definite integral quickly - though of course that could involve first evaluating the indefinite integral. $\endgroup$ – Peter Woolfitt Apr 19 '15 at 23:02


I suppose that the change of variable is $x=\sinh(y)$, $dx=\cosh(y)\,dy$ and so $$I=\int\sqrt{x+\sqrt{x^2+1}}\,dx=\int\sqrt{\sinh(y)+\cosh(y)} \cosh(y)\,dy=\int \cosh(y)\,e^{y/2}\,dy$$ $$I=\int\frac{e^y+e^{-y}}2\,e^{y/2}\,dy=\frac{1}{3} e^{3 y/2}-e^{-y/2}$$ and, back to $x$, the formula given by Wolfram Alpha.

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    $\begingroup$ This definitely seems to be the way to do it. $\endgroup$ – davidlowryduda Apr 19 '15 at 4:52
  • $\begingroup$ Nice! Using hyperbolic functions completely did not occur to me. Another one for the bag of tricks I guess. $\endgroup$ – Peter Woolfitt Apr 19 '15 at 4:57
  • $\begingroup$ @Claude Leibovici Well done! +1. $\endgroup$ – Olivier Oloa Apr 19 '15 at 6:22
  • $\begingroup$ @OlivierOloa. Be sure that, this coming from you, I really appreciate ! Thanks :-) $\endgroup$ – Claude Leibovici Apr 19 '15 at 6:23
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    $\begingroup$ Beat me to it by two whole hours ! :-$)$ $\endgroup$ – Lucian Apr 19 '15 at 6:53

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int \sqrt{x+\sqrt{x^2+1}}dx$$

Now Let $$\left(x+\sqrt{x^2+1}\right)=t^2....................\color{red}\checkmark$$

Then $$\displaystyle \frac{\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}dx = 2tdt$$

So We get $$\displaystyle dx = \frac{2\sqrt{x^2+1}}{t}dt\;, $$ Now for Calculation of $$\sqrt{x^2+1}$$ in terms of $t\;,$ We are Using

$$\displaystyle \left(x+\sqrt{x^2+1}\right)\cdot \left(x-\sqrt{x^2+1}\right) = t^2\cdot \left(x-\sqrt{x^2+1}\right)$$

So We get $$\displaystyle \left(\sqrt{x^2+1}-x\right)=\frac{1}{t^2}..................\color{red}\checkmark$$

So We Get $$\displaystyle \sqrt{x^2+1}dx = \frac{t^4+1}{2t^2}dt$$

So We get $$\displaystyle I = \int\frac{t\cdot 2(t^4+1)}{2t^3}dt = \int \left(t^2+t^{-2}\right) dt = 2t-\frac{1}{t}+\mathcal{C}$$

So We Get $$\displaystyle \int\sqrt{x+\sqrt{x^2+1}}dx = 2\sqrt{x+\sqrt{x^2+1}}+\sqrt{x-\sqrt{x^2+1}}+\mathcal{C}$$

  • $\begingroup$ Nice way ! If fact, you can have make it shorter since $\left(x+\sqrt{x^2+1}\right)=t^2$ gives $x=\frac{t^4-1}{2 t^2}$ $\endgroup$ – Claude Leibovici Apr 19 '15 at 8:29
  • $\begingroup$ Nice solution! Just a tiny correction - I think the $dx$ and $dt$ shouldn't be in the equation immediately after your second red check mark. $\endgroup$ – Peter Woolfitt Apr 19 '15 at 23:27

$$ \begin{align} \int_0^{1/\sqrt3}\sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x &=\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\mathrm{d}\tan(\theta)\tag{1}\\ &=\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{2}\\ &=\frac1{\sqrt[4]3}-\int_0^{\pi/6}\tan(\theta)\frac{\sec^2(\theta)+\tan(\theta)\sec(\theta)}{2\sqrt{\tan(\theta)+\sec(\theta)}}\,\mathrm{d}\theta\tag{3}\\ &=\frac1{\sqrt[4]3}-\frac12\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\mathrm{d}\sec(\theta)\tag{4}\\ &=\frac12+\frac12\int_0^{\pi/6}\sec(\theta)\frac{\sec^2(\theta)+\tan(\theta)\sec(\theta)}{2\sqrt{\tan(\theta)+\sec(\theta)}}\,\mathrm{d}\theta\tag{5}\\ &=\frac12+\frac14\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{6}\\[6pt] &=\frac23\tag{7} \end{align} $$ Explanation:
$(1)$: substitution $x\mapsto\tan(\theta)$
$(2)$: $\frac{\mathrm{d}}{\mathrm{d}\theta}\tan(\theta)=\sec^2(\theta)$
$(3)$: integrate by parts
$(4)$: $\frac{\mathrm{d}}{\mathrm{d}\theta}\sec(\theta)=\tan(\theta)\sec(\theta)$
$(5)$: integrate by parts
$(6)$: simplification
$(7)$: $\frac43$ of $(6)$ minus $\frac13$ of $(2)$

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    $\begingroup$ So simple ... after reading your answer ! $\endgroup$ – Claude Leibovici Apr 19 '15 at 13:37

If you have dealt with the explicit formulas for arcsinh and arccosh you might recognize that $${\rm arcsinh}\,\, x = \log (x + \sqrt{x^2 + 1})$$ So you are being asked to integrate $$\int_0^{1 \over \sqrt{3}} e^{{1 \over 2} {\rm arcsinh}\,x}\,dx$$ So you are led to substituting $x = \sinh y$ and the integral is $$\int_0^{\sinh^{-1}{1 \over \sqrt{3}}} e^{y \over 2} \cosh y \,dy$$ $$= \int_0^{\sinh^{-1} {1 \over \sqrt{3}}} {e^{3y \over 2} + e^{-{y \over 2}} \over 2} \,dy$$ This is now an easy calculus problem although the result might be annoying due to the bounds of integration here. However, $\sinh^{-1} {1 \over \sqrt{3}} = \ln ({1 \over \sqrt{3}}+ \sqrt{{1 \over 3} + 1})= \ln \sqrt{3} = {1 \over 2}\ln 3$, so the result is $$\big({1 \over 3} e^{3y \over 2} - e^{-{y \over 2}}\big)\bigg\vert_0^{{1 \over 2} \ln 3}$$ $$= {1 \over 3} e^{{3 \over 4}\ln 3} - e^{-{\ln 3 \over 4}} + {2 \over 3}$$ $$= {1 \over 3} 3^{3 \over 4} - 3^{-{1 \over 4}} + {2 \over 3}$$ $$ = {2 \over 3}$$


Also works well with $x=\frac12\left(t-\dfrac1t\right)$.

$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=\frac12\int\sqrt{\frac12\left(t-\frac1t\right)+\frac12\left(t+\frac1t\right)}\left(1+\frac1{t^2}\right)\,dt\\ =\frac12\int\left(t^{1/2}+t^{-3/2}\right)\,dt=\frac13t^{3/2}-t^{-1/2}.$$

To transform the bounds, $t^2-2xt-1=0$ i.e. $t=x+\sqrt{x^2+1}$, giving $1$ to $\sqrt3$.

  • $\begingroup$ Wow, this is a neat approach too! $\endgroup$ – Peter Woolfitt Apr 19 '15 at 23:34
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    $\begingroup$ Yep, actually it is the hyperbolic substitution, without exponential functions. The crux of the method is to somehow get rid of $\sqrt{x^2+1}$. $\endgroup$ – Yves Daoust Apr 20 '15 at 7:13

Why not just substitute the whole complicated integrand?

This seems to be the simplest approach, straighforward, no guesswork necessary.

$$u = \sqrt{\sqrt{x^2+1}+x}$$

Then we have

$$u(x=0) = 1, u(x=\frac{1}{\sqrt{3}})= \sqrt[4]{3}\tag{*}$$

$$x = \frac{u^4-1}{2 u^2}$$

$$\,dx = (\frac{1}{u^3}+u)\,du$$

And the indefinite integral is elementary

$$i = \int \left(\frac{1}{u^3}+u\right) u \, du = \frac{u^3}{3}-\frac{1}{u}$$

In the limits (*) it becomes $\frac{2}{3}$.


Using formula: $$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$ so: $$\sqrt{x+\sqrt{x^2+1}}=\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}$$ and then:

$$\int \left(\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}\right) \, dx=\frac{1}{3} \sqrt{2} (-i+x)^{3/2}+\frac{1}{3} \sqrt{2} (i+x)^{3/2}+C$$

  • $\begingroup$ The only drawback is you need to know the formula :( $\endgroup$ – Mariusz Iwaniuk Oct 25 '18 at 16:38

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