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

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- Feb 27th 2010, 05:07 PMkaiser0792Integration 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 PMGeneral
- Feb 27th 2010, 05:33 PMJhevon
Let $\displaystyle u = \sqrt x$

$\displaystyle \Rightarrow u^2 = x$

$\displaystyle \Rightarrow 2u~du = dx$

So your integral becomes

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

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

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

I suppose the rest will be easy for you - Feb 27th 2010, 05:45 PMkaiser0792Thanks 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 PMkaiser0792
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 PMGeneral
- Feb 27th 2010, 08:20 PMkaiser0792
Thanks for your help and for your follow up.