Integral of a function with a square root in denominator

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Homework Statement


integrate (xdx)/sqrt(1-x^2) on the integreal [2,5]


Homework Equations


integral u^n(du) = u^n+1/(n+1)


The Attempt at a Solution


I'm no good at using latex so I scanned in my work.

I get .716 but when I check with my calculator it comes out as 2.863
 

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I see where I messed up. I was dividing u^1/2 by 2 instead of 1/2. Thanks anyways.
 


Two things:

\int u^{-1/2} du = \frac{u^{1/2}}{-\frac{1}{2}+1} = 2 \sqrt{u}.

Also you forgot to change the limits of integration after making the change of variables.
 


fzero said:
Two things:

\int u^{-1/2} du = \frac{u^{1/2}}{-\frac{1}{2}+1} = 2 \sqrt{u}.

Also you forgot to change the limits of integration after making the change of variables.

Actually I think I did do that. Not sure if u can see on the side where i did u(2) =5 and u(5) = 26.
 
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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