# indefinite integral, including e

• Apr 14th 2009, 06:03 AM
foodmmm
indefinite integral, including e
$

\int (x^2-1)e^{x^3-3x}dx
$
• Apr 14th 2009, 06:12 AM
apcalculus
Quote:

Originally Posted by foodmmm
$

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

multiply inside the integral by 3, and divide on the outside by 3.

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

Good luck!!
• Apr 14th 2009, 06:15 AM
foodmmm
:(
what do i do with the e?
• Apr 14th 2009, 06:18 AM
apcalculus
Quote:

Originally Posted by foodmmm
what do i do with the e?

Use the Chain Rule backwards:

$\int f'(x) e^{f(x)} dx = e^{f(x)} + C$

Note that $f(x) = x^3 - 3x$ and $f'(x) = 3x^2 - 3$, so the antiderivative is simply
$e^{x^3 - 3x} + C$

Good luck!!
• Apr 14th 2009, 06:32 AM
foodmmm
confused
sorry, where did the 1/3 go?
• Apr 14th 2009, 06:34 AM
apcalculus
Quote:

Originally Posted by foodmmm
sorry, where did the 1/3 go?

ops! the 1/3 goes in the front!

so the final answer is $\frac{1}{3} e^{x^3 - 3x} + C$
• Apr 14th 2009, 06:42 AM
foodmmm
lost
so how do you get from

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

to

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

im not sure how to take those steps in between.
• Apr 14th 2009, 06:46 AM
apcalculus
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 AM
mr fantastic
Quote:

Originally Posted by foodmmm
so how do you get from

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

to

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

im not sure how to take those steps in between.

You were given the rule in post #4.

So either use that rule or make a substitution. The substution in this case is $u = x^3 - 3x$.