# Help with indefinite integrals?

• Nov 28th 2010, 11:46 AM
yess
Help with indefinite integrals?
antiderivative [(3x)/((2x^2+1)^1/2)]
what i did:
let u=2x^2+1
du/dx=4x
(3/4)du=(3/4)(4x)(dx)
3/4 du=(3x)(dx)
x=[(u-1)/2)]^1/2
antiderivative [(u^1/2)(3[(u-1)/2]^1/2)(du)]
= antiderivative [(3/u^1/2)[(u-1)/2]^1/2
from here i get stuck..i need to multiply du in as well but im not sure how to do that since i only found 3/4du=3xdx
how would i do this? or am i doing something wrong?
• Nov 28th 2010, 11:55 AM
pickslides
This looks OK to me

$\displaystyle \displaystyle \int \frac{3x}{\sqrt{2x^2+1}}~dx = \frac{3}{4}\int \frac{4x}{\sqrt{2x^2+1}}~dx =\frac{3}{4} \int \frac{1}{\sqrt{u}} = \frac{3}{4}\frac{\sqrt{u}}{\frac{1}{2}}+C=\dots$
• Nov 28th 2010, 12:13 PM
yess
Quote:

Originally Posted by pickslides
This looks OK to me

$\displaystyle \displaystyle \int \frac{3x}{\sqrt{2x^2+1}}~dx = \frac{3}{4}\int \frac{4x}{\sqrt{2x^2+1}}~dx =\frac{3}{4} \int \frac{1}{\sqrt{u}} = \frac{3}{4}\frac{\sqrt{u}}{\frac{1}{2}}+C=\dots$

Thank youu =]