e^-x x^2 dx?

Question

Integration by parts - Step by step

Answer 1

Let u=x^2 and let v = e^-x
du = 2xdx

So, you have x^2e^-x - 2INT(e^-x*x dx)

OK Let u = x and let v = e^-x => du =dx
So xe^-x-INT(e^-x dx)
xe^-x+e^-x + C

x^2e^-x + 2xe^-x +2e^-x + C //QED

Answer 2

Through integration by parts we want to reduce x^2 to 2x, then to a single number.

Therefore, first take f(x) = -e^(-x) and g(x) = x^2, then

e^(-x) x^2 = f'(x) g(x)
so INT f'(x) g(x) dx = f(x) g(x) - INT f(x) g'(x) dx gives
INT e^(-x) x^2 dx = -e^(-x) x^2 - INT [-e^(-x) 2x] dx
or
... = -e^(-x) x^2 + 2 INT e^(-x) x dx

Next, we must integrate e^(-x) x dx -- same way. Take f(x) = -e^(-x) and g(x) = x, then

e^(-x) x = f'(x) g(x)
so INT f'(x) g(x) dx = f(x) g(x) - INT f(x) g'(x) dx gives
INT e^(-x) x dx = -e^(-x) x - INT [-e^(-x) 1] dx
or
... = -e^(-x) x + INT e^(-x) dx = -e^(-x) x - e^(-x)

Combining everything, the final result is

-e^(-x) x^2 + 2 [-e^(-x) x - e^(-x)]
= -e^(-x) [x^2 + 2x + 2]

In general, INT e^(-x) x^n dx leads to -e^(-x) times a polynominal of degree n.