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Math Help - Lost on Hard Substitution Indefinite Integral

  1. #1
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    Lost on Hard Substitution Indefinite Integral

    \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!
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  2. #2
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    By differentiating x^3-3x we get 3x^2-3=3(x^2-1). What's the substitution?
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  3. #3
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    Quote Originally Posted by Jim Marnell View Post
    \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?
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  4. #4
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    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!
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  5. #5
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    Indefinite Integral Substitution Problem-Just needs checked

    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
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  6. #6
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    That's OK !
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