Hi, I'm new to integration. Can anyone help me figure out where I went wrong?

Many thanks.

Q.Evaluate the following:

Please see attachment.

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- Nov 15th 2011, 01:37 PMGrigOrig99Integration by Substitution
Hi, I'm new to integration. Can anyone help me figure out where I went wrong?

Many thanks.

**Q.**Evaluate the following:

Please see attachment. - Nov 15th 2011, 01:44 PMtom@ballooncalculusRe: Integration by Substitution
Notice your u version of the integrand has no traces of the x in the numerator of the original. To bring that part with you, you can re-express it (x, that is) in terms of u, or else use integration by parts.

- Nov 15th 2011, 01:47 PMe^(i*pi)Re: Integration by Substitution
- Nov 15th 2011, 01:47 PMQuackyRe: Integration by Substitution
You haven't taken into account the $\displaystyle x$ on the numerator.

We have:

$\displaystyle \int^4_0{\frac{x}{\sqrt{2x+1}}}dx$

I'm going to let $\displaystyle u^2=2x+1$ so that $\displaystyle 2u~du=2~dx$ and $\displaystyle dx=u~du$

When $\displaystyle x=4$, $\displaystyle u=3$ and when $\displaystyle x=0$, $\displaystyle u=1$

I have:

$\displaystyle \int^4_0{\frac{x}{\sqrt{2x+1}}}dx$

=$\displaystyle \int^3_1{\frac{u^2-1}{2u}}\cdot{u}~du$

Take it! - Nov 15th 2011, 02:07 PMtom@ballooncalculusRe: Integration by Substitution
The parts version...

http://www.ballooncalculus.org/draw/parts/thirty.png

... where (key in spoiler) ...

__Spoiler__:

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Nov 15th 2011, 02:39 PMGrigOrig99Re: Integration by Substitution
Thanks guys.