`int xe^(-4x) dx` Find the indefinite integral

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Given

`int x(e^(-4x)) dx`

by applying  integration by parts, we'll get the answer

let `u=x => u'= 1`

`v'=e^(-4x) so , v= -1/4e^(-4x)`

Now by integration by parts ,

`int uv' dx = uv - int u'v dx`

so ,

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

=`-x/4e^(-4x)...

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Given

`int x(e^(-4x)) dx`

by applying  integration by parts, we'll get the answer

let `u=x => u'= 1`

`v'=e^(-4x) so , v= -1/4e^(-4x)`

Now by integration by parts ,

`int uv' dx = uv - int u'v dx`

so ,

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

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

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

let us find

`int e^(-4x) dx`

let `u= -4x`

`du = -4dx`  so `dx = -1/4du`

so,

`int e^(-4x) dx= int e^(u) -1/4du`

=`-1/4int e^u du`

=`-1/4e^u = -1/4e^(-4x)`

so, now

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

=`-x/4e^(-4x) +1/4 (-1/4)e^(-4x)`

=`-x/4e^(-4x) -1/16e^(-4x) +C`

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