Question

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

Answer 1

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 ^_^

Answer 2

substitute and solve :)

Answer 3

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