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
No problem! Mistakes happen, and it's great that you caught it. Let's correct it together.
Starting with your substitution u = sqrt(x) + 1, we have:
sqrt(x) = u - 1
x = (u - 1)^2
dx = 2(u - 1) du
Substituting these back into the integral, we get:
∫(1 / (sqrt(x) + 1)^2) dx
= ∫(1 / (u)^2) (2(u - 1) du)
= ∫(2(u - 1) / u^2) du
= 2∫((u - 1) / u^2) du
Now we can split this integral into two parts:
2∫(u/u^2) du - 2∫(1/u^2) du
The first integral simplifies to:
2∫(1/u) du = 2 ln|u| + C
The second integral simplifies to:
-2∫(1/u^2) du = -2u^(-1) - 2C = -2/u - 2C
Putting it all together, we get:
2 ln|u| - 2/u + C
Finally, substitute back u = sqrt(x) + 1:
2 ln|sqrt(x) + 1| - 2/(sqrt(x) + 1) + C
And there you have it! The integral of (sqrt(x) + 1)^2 is 2 ln|sqrt(x) + 1| - 2/(sqrt(x) + 1) + C.