Calculus URGENT test tonight
posted by Gabriela .
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 = u1
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
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