>>927806>Integration by partsThis is a technique used to deal with an integral of a product of two functions, one that can be endlessly derived/integrated, and one that can be gone within a finite steps of derivation.
In your assignment, you have an integral of x^3 times e^-3x. You know that e^ax can be integrated/derived repeatedly without end. You should also know that x^3 can be derived 3 times before it becomes a constant. So now look at pic related. You need to define f(x) as the finite function, because on the right side you can see it gets derived once, and if you do this procedure enough times, the repeated derivation will get rid of the finite function. You then need to define g'(x) as the infinite function, the exponent or in some cases a sine/cosine. g(x) is then the integral of the infinite function. You should generally know that the integral of e^ax is (e^ax)/a, when "a" is a constant". In this case, a=-3. f'(x) is of course the derivative of the finite function. The derivative of x^a is a*x^(a-1), for "a" being a constant.
All you have left to do is plug in the functions into the formula. Once you do so, you'll have a non-integrated part and an integrated part. Ignore the first part and focus on the new integral. Now the power of x should be reduced to 2. Feel free to push constants out of the integral. Use the formula again, but reassign f(x) and g'(x) to the functions of the new integral, one being x^2, the other being e^-3x, and again detail g(x) and f'(x). By using the formula, you'll have a second non-integrated expression and a new integral featuring x*e^-3x. Use the integral by parts one more time, and you'll see that x disappears from the last integral. All you have left is e^-3x. Integrate that silly goose like you did 3 times already, and you should have 4 or so expressions featuring some constant times some power of x times e^-3x. You can take e^-3x out as a common denominator, and you're done.
It's long, but it's not a hard process.
