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Problem

Find $$\int \frac{\arcsin \sqrt{x}}{(1+x)\sqrt{x}}{\rm d}x.$$

My Try

Considering making a substitute, let $\arcsin\sqrt{x}=t$, where $0<t\leq \dfrac{\pi}{2}.$ Then we obtain $$x=\sin^2 t,{\rm d}x=2\sin t \cos t{\rm d}t.$$Thus,$$\int \frac{\arcsin \sqrt{x}}{(1+x)\sqrt{x}}{\rm d}x=2\int \frac{t\cos t}{1+\sin^2 t}{\rm d}t.$$

But I'm stuck here. How to go on with this?

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mengdie1982
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    My "spidey sense" is telling me that there is probably no way to represent your integral in terms of elementary functions. – Frank W May 21 '18 at 01:14
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    Also have you tried letting $u=\sqrt{x}$ and then using partial fractions? That might simplify your work a bit, I'm not sure though... – Frank W May 21 '18 at 01:15
  • http://www.wolframalpha.com/input/?i=integral+arcsin(sqrt(x))%2F((1%2Bx)(sqrt(x))) if you just made this up don't do that –  May 21 '18 at 01:18
  • @ZacharySelk It's either my wifi, or the link, because Wolfram Alpha just keeps on loading... – Frank W May 21 '18 at 01:19
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    The link is fine. It takes a while to compute. – Randall May 21 '18 at 01:19
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    I'm voting to close this question as off-topic because the integral has no nice expression. –  May 21 '18 at 01:21
  • The substitution $u=\sqrt{x}$ obtains an integral I discussed in https://math.stackexchange.com/questions/2778024/first-step-for-int-frac-operatornamegd-left-log-x-right-sqrt1-x2dx Unfortunately, as others have noted, it lacks a nice closed form. – J.G. Jun 13 '18 at 15:33
  • Simplifying: $\text{Li}_2\left(\left(-1-\sqrt{2}\right) e^{i \sin ^{-1}\left(\sqrt{x}\right)}\right)-\text{Li}_2\left(\left(1-\sqrt{2}\right) e^{i \sin ^{-1}\left(\sqrt{x}\right)}\right)+\text{Li}_2\left(\left(-1+\sqrt{2}\right) e^{i \sin ^{-1}\left(\sqrt{x}\right)}\right)-\text{Li}_2\left(\left(1+\sqrt{2}\right) e^{i \sin ^{-1}\left(\sqrt{x}\right)}\right)-2 i \sin ^{-1}\left(\sqrt{x}\right) \left(\tanh ^{-1}\left(\left(\sqrt{2}-1\right) e^{i \sin ^{-1}\left(\sqrt{x}\right)}\right)-\tanh ^{-1}\left(\left(1+\sqrt{2}\right) e^{i \sin ^{-1}\left(\sqrt{x}\right)}\right)\right)$ – Mariusz Iwaniuk Jun 13 '18 at 16:27

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