$\displaystyle

\int (x^2-1)e^{x^3-3x}dx

$

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- Apr 14th 2009, 06:03 AMfoodmmmindefinite integral, including e
$\displaystyle

\int (x^2-1)e^{x^3-3x}dx

$ - Apr 14th 2009, 06:12 AMapcalculus
- Apr 14th 2009, 06:15 AMfoodmmm:(
what do i do with the e?

- Apr 14th 2009, 06:18 AMapcalculus
- Apr 14th 2009, 06:32 AMfoodmmmconfused
sorry, where did the 1/3 go?

- Apr 14th 2009, 06:34 AMapcalculus
- Apr 14th 2009, 06:42 AMfoodmmmlost
so how do you get from

$\displaystyle

\frac{1}{3} \int (3x^2 - 3) e^{x^3 -3x}

$

to

$\displaystyle

\frac{1}{3} e^{x^3 - 3x} + C

$

im not sure how to take those steps in between. - Apr 14th 2009, 06:46 AMapcalculus
It's the Chain Rule for Integrals. If the fully derived function is inside the integral, you can take the antiderivative easily. See the the general property I listed above in the case of the exponential function with an exponent that is also a function. As long as the first derivative is inside the integral (or can be tweaked by multiplying or diving by constant as we did), then it's very simple. The antiderivative is simply the exponential function itself, plus a constant.

Hope this helps! Good luck!! - Apr 14th 2009, 06:47 AMmr fantastic