# Calculus URGENT test tonight

posted by
**Gabriela** on
.

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