Integral of:

__1__
(sqrt(x)+1)^2 dx

The answer is:

2ln abs(1+sqrt(x)) + 2(1+sqrt(X))^-1 +c

I have no clue why that is! Please help.

I used substitution and made u= sqrt(x)+1

but i don't know what happened along the way!

Your first step was a good one.
next, let u = sqrt x + 1
Then sqrt x = u-1
x = u^2 - u + 1
dx = 2u du - du
Therefore the integral becomes
(Integral of) (2/u) du - 2/u^2 du
= 2 ln u + 2/u +c
Finally, substitute sqrt x +1 for u

When you use substitution, both the integrand (f(x)) and the differential (dx) must be changed.

thank you so very very much!

I made a mistake with two lines, bit it came out OK.
I should have written
x = u^2 - 2u + 1
dx = 2u du - 2 du