# Find the value of a hard integral [closed]

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

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

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 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.)? 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 $u = Sinh[v].$