Question

Evaluate the indefinite integral:

∫ from 1 to 2 of ln (x²e^x) dx

Thank you for your help!!

Answer 1

Since powers of decrease when you take the derivative, let

u = x² so du = 2xdx
dv = e^xx dx so v = e^xx

∫ u dv = uv = ∫ v du
= x²e^x - ∫2xe^x dx

Now let u = x so du = dx
dv = e^xdx so v = e^x

∫2xe^x dx = xe^x - ∫e^x dx = xe^x - e^x

So, ∫x²e^x = x²e^x - 2(xe^x - e^x)
= e^x(x² - 2x + 2)

Answer 2

I see there were some typos. In correct form,

u = x² so du = 2xdx
dv = e^x dx so v = e^x

∫ u dv = uv = ∫ v du
= x²e^x - ∫2xe^x dx

Now let u = x so du = dx
dv = e^xdx so v = e^x

∫xe^x dx = xe^x - ∫e^x dx = xe^x - e^x

So, ∫x²e^x = x²e^x - 2(xe^x - e^x)
= e^x(x² - 2x + 2)