Integral of xe^x dx with substution please?

u = x . . . . . dv = e^x dx

du = dx . . . . v = e^x

∫ u dv = u v - ∫ v du

∫ x e^x dx = x e^x - ∫ e^x dx

∫ x e^x dx = x e^x - e^x + C

Mαthmφm

you can use substitution but you are going to end up using integration by parts anyways.

u=x

du=1

dv=e^x

v=e^x

xe^x-integral[e^x]

xe^x-e^x+C.

with substituion.

t=x

dx=dt

integral of te^tdt. As you see here this is the same form as xe^x. And x=t so using substitution doesn't really make this problem easier.

Let u = x; du = dx and dv = e^xdx; v = e^x

Then use ∫udv = uv-∫vdu

(x-1)(e^x)