# Thread: indefinite integral, including e

1. ## indefinite integral, including e

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

2. Originally Posted by foodmmm
$\displaystyle \int (x^2-1)e^{x^3-3x}dx$

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

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

Good luck!!

3. ## :(

what do i do with the e?

4. Originally Posted by foodmmm
what do i do with the e?
Use the Chain Rule backwards:

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

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

Good luck!!

5. ## confused

sorry, where did the 1/3 go?

6. Originally Posted by foodmmm
sorry, where did the 1/3 go?
ops! the 1/3 goes in the front!

so the final answer is $\displaystyle \frac{1}{3} e^{x^3 - 3x} + C$

7. ## lost

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.

8. 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!!

9. Originally Posted by foodmmm
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.
You were given the rule in post #4.

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