- #1

- 29

- 0

soln'.... let u=1-x... then -du=dx and x^2 = (1-u)^2 (sub back in)...

this gives me -2*u^(1/2)*(15-10*u+3*u^2) /15.... (sorry for lack of proper terms)

anyway, this turns out to be wrong...where did i go wrong here?

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- Thread starter johnnyboy2005
- Start date

- #1

- 29

- 0

soln'.... let u=1-x... then -du=dx and x^2 = (1-u)^2 (sub back in)...

this gives me -2*u^(1/2)*(15-10*u+3*u^2) /15.... (sorry for lack of proper terms)

anyway, this turns out to be wrong...where did i go wrong here?

- #2

benorin

Homework Helper

- 1,344

- 136

[tex]\int \frac{x^2}{\sqrt{1-x}}dx = -\int \frac{(1-u)^2}{\sqrt{u}}du = -\int u^{-\frac{1}{2}}(1-2u+u^2) du[/tex]

then distribute the [itex]u^{-\frac{1}{2}}[/itex] term over the quadratic and integrate

- #3

- 29

- 0

yeah...that's what i was getting but it was in a different form. Thanks a lot !!1

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