# Use integration by parts to integrate Integration sign, x^2e^2x dx e is to the power of 2x

beckden | High School Teacher | (Level 1) Educator

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Use LIPET to get `u = x^2` , then `dv = e^(2x) dx`

This give `du = 2x dx` and `v = 1/2 e^(2x)`

Use `int u dv = uv - int v du`

`int x^2 e^(2x) dx =x^2 (1/2)e^(2x) - int (1/2 e^(2x)) 2x dx`

`int x^2 e^(2x) dx = 1/2x^2e^(2x) - int x e^(2x) dx`

We now use u = x and `dv = e^(2x)` and get `du = dx` and `v = 1/2e^(2x)`

Integrating by parts again to get

` int x^2 e^(2x) dx = 1/2x^2e^(2x) - (x(1/2e^(2x)) - int 1/2e^(2x) dx)`

`int x^2 e^(2x) dx = 1/2x^2e^(2x) - 1/2xe^(2x) + 1/4 e^(2x) + C`

`int x^2 e^(2x) dx = 1/2e^(2x)(x^2 - x + 1/2) + C`

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

using integration by parts

`int udv = uv-intvdu`

Let u=x^2 and v=e^(2x)

Then ;

du = 2xdx

dv = 2e^(2x).dx

`int(x^2*2e^(2x))dx`

`= (x^2)(e^(2x))-int(e^(2x)*2x)dx`

In the same manner by taking u=2x and v=e^2x you can get;

`int(e^(2x)*2x)dx` `= 2[x/2-1/4]*e^2x`

`int(x^2*2e^(2x))dx`

= `(x^2)(e^(2x))`-`2[x/2-1/4]*e^(2x)` +C

Therefore;

`int(x^2*e^(2x))dx=(1/2)((x^2)(e^(2x)))-2[(x/2-1/4)*e^(2x)]+C`