Integration

**Tangera** · September 13th 2008, 08:25 AM

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!

**Moo** · September 13th 2008, 08:27 AM

Hello !

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 ^^

---------------------------------------
Otherwise, you can do it easily by parts

$u=x \implies du=dx$

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

Thread: Integration

LinkBack

Thread Tools

Display

Integration

Similar Math Help Forum Discussions

Integration help please these are three small even basic integration problems

sketch the region of integration and change the order of integration.

Definite Integration - Double integration by parts

Simple Integration problem: Trigonometric Integration

Tricky integration, possibly Integration Factor

Search Tags