Am I doing this integration problem right?

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Homework Statement


Hello, I am struggling with an integration problem.

e ^ x ^ (5/3)

Homework Equations



Um.

The Attempt at a Solution



So I know the integral of x^(5/3) is (3/8)x^(8/3). But when I am trying to integrate the whole thing, with e in it, do I just treat x^(5/3) like a constant? Or do I still have to integrate it?

Possible solutions: e^x^(5/3)
or
e^(3/8)x^(8/3)
or
something else... please help...
 
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Take a close look at the problem. Does it appear to be (e^x)^(5/3) or is it e^(x^(5/3))?

The first is relatively easy (simplify and substitute). If in doubt, I'd assume this interpretation.

The second is really, really ugly. I can't figure out how to do it, my TI-89 chokes on it, and Maple spits out the following gibberish:
output.png

so I'm guessing it's not really solvable by hand using first/second-year calculus techniques.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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