I think this shouldn't be difficult but I can't figure it out! >.<

$\displaystyle \int\displaystyle{\frac{2x}{\sqrt{x+1}}}dx$

Thank you for helping!

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- Sep 13th 2008, 08:25 AMTangeraIntegration
I think this shouldn't be difficult but I can't figure it out! >.<

$\displaystyle \int\displaystyle{\frac{2x}{\sqrt{x+1}}}dx$

Thank you for helping! - Sep 13th 2008, 08:27 AMMoo
Hello !

Substitute $\displaystyle u=\sqrt{x+1}$

$\displaystyle \frac{du}{dx}=\frac{1}{2 \sqrt{x+1}}=\frac 1{2u}$

Thus $\displaystyle dx=2u du$

$\displaystyle \int \frac{2x}{\sqrt{x+1}} ~dx=\int \frac{2x}{u} \times (2u) ~du$

and $\displaystyle x+1=u^2 \implies x=u^2-1$

Finish it ^^

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Otherwise, you can do it easily by parts (Wink)

$\displaystyle u=x \implies du=dx$

$\displaystyle v'=\frac{dx}{\sqrt{x+1}} \implies v=2 \sqrt{x+1}$