# Thread: Lost on Hard Substitution Indefinite Integral

1. ## Lost on Hard Substitution Indefinite Integral

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

dont know any rules for these type of problems involving e. Thanks for any help in getting me started with this problem!

2. By differentiating $\displaystyle x^3-3x$ we get $\displaystyle 3x^2-3=3(x^2-1).$ What's the substitution?

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

dont know any rules for these type of problems involving e. Thanks for any help in getting me started with this problem!
you want u and du to substitute for everything inside the integral.
1. You can do this by setting u= x^3-3x which gives du=3(x^2-1)dx get rid of the 3 by dividing both sides by it and you'll get 1/3du= (x^2-1)dx which is what you want.
2. Substitute into integral:
should get something like this by taking the 1/3 out as a constant
1/3 integral e^u du , which shows everything is accounted for from the original.
3. Integrate. When you do you should get 1/3 e^u + C
4. Plug u back in and you're done.

1/3 e^(x^3-3x) + C

Is it clear now?

4. Thank you! yea it makes sense, i had what you had in step 1 and then everything else i had was wrong but i'll follow what you did for my other questions like this. Thanks again!

5. ## Indefinite Integral Substitution Problem-Just needs checked

Just needs checked: Not sure if i did it right

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

$\displaystyle u=x^3-3x$

$\displaystyle du=3x^2-3$

$\displaystyle du=3(x^2-1)dx$

$\displaystyle \frac{du}{3}=(x^2-1)dx$

$\displaystyle \frac{1}{3}du=(x^2-1)dx$

$\displaystyle \frac{1}{3}e^u(du)$

$\displaystyle \frac{1}{3}e^u+c$

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

6. That's OK !