# Integration Proglem

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• Feb 27th 2010, 05:07 PM
kaiser0792
Integration Proglem
I'm trying to integrate the function 1/(1 + x^1/2) dx. Any help would be greatly appreciated.
• Feb 27th 2010, 05:33 PM
General
Quote:

Originally Posted by kaiser0792
I'm trying to integrate the function 1/(1 + x^1/2) dx. Any help would be greatly appreciated.

Let $u=\sqrt{x}$ to get:

$2\int \frac{u}{u+1} du$.
Can you complete it?
• Feb 27th 2010, 05:33 PM
Jhevon
Quote:

Originally Posted by kaiser0792
I'm trying to integrate the function 1/(1 + x^1/2) dx. Any help would be greatly appreciated.

Let $u = \sqrt x$

$\Rightarrow u^2 = x$

$\Rightarrow 2u~du = dx$

So your integral becomes

$2 \int \frac u{1 + u}~du$

$= 2 \int \left( \frac {u + 1 - 1}{u + 1} \right)~du$

$= 2 \int \left( 1 - \frac 1{u + 1} \right)~du$

I suppose the rest will be easy for you
• Feb 27th 2010, 05:45 PM
kaiser0792
Thanks General
That did it! I missed squaring u and adding and subtracting 1.

I really appreciate the help, it was driving me crazy!!
• Feb 27th 2010, 06:11 PM
kaiser0792
let u = x^1/2, u^2 = x and then 2u du = dx.

Rewriting: 2*integral (u/u+1) du = 2*integral [u+1-1]/(u+1)du
= 2[integral (u+1)/(u+1)du - integral 1/(u+1)du]
= 2[integral du - integral 1/(u+1)du] Let v = u+1, dv=du
= 2[u - integral dv/v = 2u - ln|v|] + C
= 2[x^1/2 - ln|u+1|] + C = 2[x^1/2 - ln|(x^1/2 + 1)|] + C
• Feb 27th 2010, 07:49 PM
General
Quote:

Originally Posted by kaiser0792
let u = x^1/2, u^2 = x and then 2u du = dx.

Rewriting: 2*integral (u/u+1) du = 2*integral [u+1-1]/(u+1)du
= 2[integral (u+1)/(u+1)du - integral 1/(u+1)du]
= 2[integral du - integral 1/(u+1)du] Let v = u+1, dv=du
= 2[u - integral dv/v = 2u - ln|v|] + C
= 2[x^1/2 - ln|u+1|] + C = 2[x^1/2 - ln|(x^1/2 + 1)|] + C

Its correct. :)
• Feb 27th 2010, 08:20 PM
kaiser0792
Thanks for your help and for your follow up.