How do you intergrate (x^2-1)e^xdx?

Answer 1

You'll have to use integration by parts. The basic formula is
integral(u dv) = u v - integral(v du)

In this case, you start by setting u = x^2 - 1, dv = e^x dx
then you can differentiate/integrate to get du = 2x dx, v = e^x

integral[(x^2-1)e^x dx] = (x^2 - 1)e^x - integral(e^x * 2x * dx)
= (x^2-1)e^x - 2 * integral(x e^x dx)

You need to repeat integration by parts again, this time with u = x and dv = e^x dx, which gives du = dx and v = e^x

= (x^2 - 1) e^x - 2 * [x e^x - integral(e^x dx)]
= (x^2 - 1) e^x - 2x e^x + 2 e^x + C
which simplifies to
x^2 e^x - 2x e^x + e^x + C

Answer 2

Separate it into two integrals: The first integrand is x^2 * e^x and the second is e^x. Use integration by parts to interate x^2 * e^x, preferably the tabular form, since it's quicker.

Answer 3

By parts: let u = x^2 - 1, dv = e^x dx
∫u dv = uv - ∫v du
So ∫(x^2 - 1) e^x dx = (x^2 - 1) e^x - ∫ e^x (2x) dx
Integrate by parts again:
= (x^2 - 1) e^x - (2x e^x - ∫ e^x (2) dx)
= (x^2 - 1) e^x - 2x e^x + 4e^x + c
= e^x (x^2 - 2x + 3) + c.

Answer 4

using integration by part
the answer is (X^2-1)e^x - 2xe^x + 2e^x
simplify it yourself