$\displaystyle \int ((x^2-1)e^{x^3-3x})dx$
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?
Just needs checked: Not sure if i did it right
$\displaystyle \int (x^2-1)(e^{x^3-3x})dx$
$\displaystyle u=x^3-3x$
$\displaystyle du=3x^2-3$
$\displaystyle du=3(x^2-1)dx$
$\displaystyle \frac{du}{3}=(x^2-1)dx$
$\displaystyle \frac{1}{3}du=(x^2-1)dx$
$\displaystyle \frac{1}{3}e^u(du)$
$\displaystyle \frac{1}{3}e^u+c$
$\displaystyle \frac{1}{3}e^{x^3-3x}+c$