Calculus Problem Definite Integral from 0 to 1 of (x^2+1)e^-x dx?

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Hey guys doing some practice problems for homework could use a little help on this one to check my answer, steps would help a ton as im having trouble simplifying.

The problem is

Definite Integral from 0 to 1 of (x^2+1)e^-x dx

captain-matticus-landpiratesinc

e^(-x) * dx + x^2 * e^(-x) * dx

-e^(-x) + int(x^2 * e^(-x) * dx)

u = x^2

du = 2x * dx

dv = e^(-x) * dx

v = -e^(-x)

-e^(-x) + (x^2 * (-e^(-x)) - (-2) * int(x * e^(-x) * dx)) =>

-e^(-x) - e^(-x) * x^2 + 2 * int(x * e^(-x) * dx)

u = x

du = dx

dv = e^(-x) * dx

v = -e^(-x) * dx

-e^(-x) - x^2 * e^(-x) + 2 * (-x * e^(-x) + int(e^(-x) * dx)) =>

-e^(-x) - x^2 * e^(-x) - 2x * e^(-x) - 2 * e^(-x) + C =>

-e^(-x) * (1 + x^2 + 2x + 2) + C =>

-e^(-x) * (x^2 + 2x + 3) + C

From 0 to 1

-e^(-1) * (1 + 2 + 3) + e^(0) * (0 + 0 + 3) =>

(-1/e) * (6) + 1 * 3 =>

3 - 6/e =>

(3e - 6) / e =>

(3/e) * (e - 2)

anonymous

Do a u du substitution. enable u = lnx. Then du = (a million/x)dx So, with changing the bounds into words of u, the quintessential is now i nt(u^2du) from ln12 to ln16. You get (u^3)/3 from ln12 to ln16. So your answer is [(ln16)^3]/3-[(ln12)^3]/3