Lost on Hard Substitution Indefinite Integral

**Jim Marnell** · April 11th 2009, 08:33 PM

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

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

**Krizalid** · April 11th 2009, 08:42 PM

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

**swimmergirl** · April 11th 2009, 08:43 PM

Originally Posted by Jim Marnell

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

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?

**Jim Marnell** · April 11th 2009, 08:46 PM

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!

**Jim Marnell** · April 14th 2009, 08:32 AM

Just needs checked: Not sure if i did it right

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

$u=x^3-3x$

$du=3x^2-3$

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

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

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

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

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

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

**running-gag** · April 14th 2009, 08:36 AM

That's OK !

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