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