Solving an Integral: Is My Attempt Right?

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Homework Statement



\int x/2x-20)^(1/2)

The Attempt at a Solution



I did u sub got the \int (u+20)/(2u^1/2)du which became (1/6)(2x-20)^(3/2) + 10(2x-20)^(1/2)

Is this right? What am i doing wrong if it is not.
 
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Do you mean
\int\frac{u+20}{2u^{1/2}}du
If so that is the same as
\int [(1/2)u^{1/2}+ 10u^{-1/2}]du
I don't see how you would get 2x-20- since there is no x in the original integral!
 
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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