Integration

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

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

Thank you for helping!
• Sep 13th 2008, 08:27 AM
Moo
Hello !
Quote:

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

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

Thank you for helping!

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

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

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

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

Finish it ^^

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

$u=x \implies du=dx$

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