how to solve the definite integral

(4x)/(x^2 -8x +16)

using integration by parts

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f(x)=4x/(x-4)^2

u=4x -> du = 4dx

dv = 1/(x-4)^2 -> v = -1/(x-4)

intf(x)=uv-intvdu

=-4x/(x-4)-int-4/(x-4)dx

=-4x/(x-4)+4ln(x-4) +C

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`f(x)=(4x)/(x-4)^2`

`u=4x -gt du = 4dx`

`dv = 1/(x-4)^2 -gt v = -1/(x-4)`

`intf(x)=uv-intvdu`

`=-(4x)/(x-4)-int-4/(x-4)dx`

`=-(4x)/(x-4)+4ln(x-4) +C`

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