Thread: Integration Proglem

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

Originally Posted by kaiser0792

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

Let to get:

.
Can you complete it?

Originally Posted by kaiser0792

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

Let





So your integral becomes







I suppose the rest will be easy for you

That did it! I missed squaring u and adding and subtracting 1.

I really appreciate the help, it was driving me crazy!!

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

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.

Thanks for your help and for your follow up.