Find the value of: $$I=\int_0^1\frac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}} \, dx$$

  • 1
    $\begingroup$ $= sinh^{-1}(e)-sinh^{-1}(1) \approx 0.844009$ See: wolframalpha.com/input/… $\endgroup$
    – jrast
    Jan 2, 2014 at 16:26
  • $\begingroup$ Are you sure the integrand is not $$\frac{\sqrt{e^x}}{\sqrt{e^x}+\sqrt{e^{-x}}}$$? If yes, the use $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ $\endgroup$ Jan 2, 2014 at 16:28
  • $\begingroup$ @labbhattacharjee if that were the integral, why not just $u=\exp x/2$ substitution? $\endgroup$
    – Igor Rivin
    Jan 2, 2014 at 16:31
  • $\begingroup$ @IgorRivin, then I think the formula I mentioned in the last comment, is more useful $\endgroup$ Jan 2, 2014 at 16:33

2 Answers 2


Making the change of variables $ u = \rm e^{x} $ gives an easier integral

$$ \int_{1}^{\rm e}\!\frac {1}{\sqrt {u^2+1}}{du}. $$

I think you can finish it now.

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    $\begingroup$ I can't help wondering why you went to so much effort to set "e" in a non-italic font in $\displaystyle\int_1^{\rm e}$ and not in the first line of your answer. Also why write {{{\rm e}}} when {\rm e} suffices (I actually changed that.)? $\endgroup$ Jan 2, 2014 at 16:43

Substituting $u = \exp(x)$ gives a standard algebraic integral: $$\int_1^e \frac{1}{\sqrt{u^2+1}} d u,$$ which falls to the standard substitution $u = Sinh[v].$


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