Integral of Exponential Help (1 Viewer)

[Forbidden.]

Find

This one in particular took "exponentially" longer than usual than all other questions.
Using du = 2xe^x dx when u = x²e^x does not yield the right answer.

 

[jyu]

Find

This one in particular took "exponentially" longer than usual than all other questions.
Using du = 2xe^x dx when u = x²e^x does not yield the right answer.

du = (2x+x²)e^x dx when u = x²e^x

 

[LoneShadow]

Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + Integral[x^2*Exp(x)dx]

let dv = Exp(x) dx => v = Exp(x)
let u = x^2 => du=2xdx

Then using Integration by parts: Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + x^2*Exp(x) - Integral[2x*Exp(x)dx] = x^2*Exp(x) + C

ooopse. I just noticesd this is in 2 unit maths section. Use jyu's method.

 

[Forbidden.]

du = (2x+x²)e^x dx when u = x²e^x

I must have differentiated u improperly.
After finding du/dx then ...

= ∫ 2xe^x+x²e^x dx
= ∫ (2x+x²)e^x dx
= ∫ du
= ∫ 1 du
= u + C
= x²e^x + C

Right ? Because it is really the working out I'm after.

Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + Integral[x^2*Exp(x)dx]

let dv = Exp(x) dx => v = Exp(x)
let u = x^2 => du=2xdx

Then using Integration by parts: Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + x^2*Exp(x) - Integral[2x*Exp(x)dx] = x^2*Exp(x) + C

ooopse. I just noticesd this is in 2 unit maths section. Use jyu's method.

Is using 2 substitutions the Extension 2 method ?
I've formatted your working out so I can comprehend it more easily.

∫[x(2+x)e^xdx] = ∫[2xe^xdx] + ∫[x²e^xdx]

let dv = e^x dx => v = e^x
let u = x² => du = 2x dx

Then using Integration by parts:
∫[x(2+x)e^xdx]
= ∫[2xe^xdx] + x²e^x - ∫[2xe^xdx] = x²e^x + C


jyu's one is a little more simpler to workout but thanks anyway.

 

[LoneShadow]

It's called "integration by parts". Pretty useful thing.

edit: and yes it's ext2. Very easy to learn though.