Posted by **Gabriela** on Friday, April 20, 2007 at 3:59pm.

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

## Answer This Question

## Related Questions

- Inequality - When I solve the inquality 2x^2 - 6 < 0, I get x < + or - ...
- Calculus - so we are doing integrals and I have this question on my assignment ...
- Math Help please!! - Could someone show me how to solve these problems step by ...
- Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...
- Calculus - Evaluate the indefinite integral: 8x-x^2. I got this but I the ...
- math calculus please help! - l = lim as x approaches 0 of x/(the square root of...
- Math(Roots) - sqrt(24) *I don't really get this stuff.Can somebody please help ...
- Math - How do you find a square root of a number that's not a perfect square? I'...
- Mathematics - sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So...
- Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...

More Related Questions