Calc2 HW Help: Solve Integral (2 + 2x - x^2)^{3/2} dx

[nitroracer]

Homework Statement

Solve the Integral
$$\int (2 + 2x - x^2)^{3/2} dx$$

2. The attempt at a solution
I have looked at integration tables, u-substitution, and integration by parts but none of the above seem to be working.

U-Substitution wouldn't work because the du would be inside the integrand.
I tried parts with U=$$(2 + 2x - x^2)^{3/2}$$ and DV=dx , but that result brought me back to the original integral.

What should I be using here that I am missing?

 

[EnumaElish]

What if you separated out 2 + 2x - x^2 and sqrt(2 + 2x - x^2), then tried integ. by parts?

 

[dynamicsolo]

U-Substitution wouldn't work because the du would be inside the integrand.
I tried parts with U=$$(2 + 2x - x^2)^{3/2}$$ and DV=dx , but that result brought me back to the original integral.

Have you had trigonometric substitutions yet? (Square roots or fractional exponents with 2 in the denominator are usually a sign that you'll want them.)

If you have, you'll need to do a couple of layers of substitution. You'd start by completing the square under the radical. You'll get a difference of two squares raised to the 3/2 power, or 1/2 power cubed. You'd then do a u-substitution, then the appropriate trig substitution. (It ain't going to be pretty, but integrating fractional powers of quadratic polynomials generally aren't...)

I should comment that integration by parts should only be considered if the "v du" integral you're going to get is something you actually know how to integrate (and preferably should be easier than the integral you began with -- or at worst, equally difficult, but related)...