e^-2.5xdx .... i got -5/2e^2,5x+c

Xcosx(3x^2)dx.... i got 1/12sin(3x^2) + c

(5x/x^2+1)dx

X(x^2-8)^1/2 dx

(1-e^-x/e^-x)dx

(3x-2)^5/2 dx

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- May 8th 2008, 03:31 PMBigBlackintegrals Help please
e^-2.5xdx .... i got -5/2e^2,5x+c

Xcosx(3x^2)dx.... i got 1/12sin(3x^2) + c

(5x/x^2+1)dx

X(x^2-8)^1/2 dx

(1-e^-x/e^-x)dx

(3x-2)^5/2 dx - May 8th 2008, 04:24 PMo_O
(Yes)

I'm guessing you mean: $\displaystyle \int x\cos \left(3x^{2}\right) dx$

Check over your work again. The constant in front of your antiderivative is off. Show your work and we'll see if we can sort it out for you.

$\displaystyle u = x^{2} + 1$

$\displaystyle u = x^{2} - 8$

Split the fraction: $\displaystyle \frac{1-e^{-x}}{e^{-x}} = \frac{1}{e^{-x}} - \frac{e^{-x}}{e^{-x}}$

And simplify.

$\displaystyle u = 3x - 2$

It'd be more helpful if you posted your work so we can pinpoint your mistakes. - May 8th 2008, 06:16 PMBigBlack
Sorry I'm new still trying to figure things out. Thanks though