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## anonymous 5 years ago I need to evaluate this integral..... x/(2x-1)^1/2 with an upper =5 and the lower =1. Can anyone Help?

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1. anonymous

let's re-write it in a better form first: $\int\limits_{1}^{5}x/\sqrt{2x-1}dx$ where you can take : $f' =({2x-1} )^{-1/2}$ g = x so now you must find g' and f in this case :) you can take the sqrt of (2x-1) up and solve using the following rule : $\int\limits_{}^{}f'gdx = fg - \int\limits_{}^{}fg'dx$ give it a try ^_^

2. anonymous

substitute and solve :)

3. anonymous

How do you take the antiderivative of (2x-1)^-1/2???

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