How do you integrate $int x^2/e^(2x)$ by integration by parts method?

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Answer 1 · 2017-02-18T12:31:07

The answer is $=-((2x^2+2x+1))/4e^(-2x)+C$

$intu'vdx=uv-intuv'dx$

We must calculate $intx^2/e^(2x)dx=intx^2e^(-2x)dx$

Let $u'=e^(-2x)$ , $=>$ , $u=e^(-2x)/(-2)=-1/2e^(-2x)$

and $v=x^2$ , $=>$ , $v'=2x$

Therefore,

$intx^2/e^(2x)dx=-1/2x^2e^(-2x)-int-1/2*2x*e^(-2x)dx$

$=-1/2x^2e^(-2x)+intxe^(-2x)dx$

In order to calculate $intxe^(-2x)dx$ , we do the integration by parts a second time

Let $u'=e^(-2x)$ , $=>$ , $u=e^(-2x)/(-2)=-1/2e^(-2x)$

and $v=x$ , $=>$ , $v'=1$

So,

$intxe^(-2x)dx=-1/2xe^(-2x)-int-1/2e^(-2x)dx$

$=-1/2xe^(-2x)+1/2inte^(-2x)dx$

$=-1/2xe^(-2x)+1/2*e^(-2x)/(-2)$

$=-1/2xe^(-2x)-1/4e^(-2x)$

Putting it all together

$intx^2/e^(2x)dx=-1/2x^2e^(-2x)-1/2xe^(-2x)-1/4e^(-2x)+C$

$=-((2x^2+2x+1))/4e^(-2x)+C$

How do you integrate int x^2/e^(2x)int x^2/e^(2x) by integration by parts method?