∫4xexdx
Apply linearity
4∫xexdx
Now solving ∫xexdx
Integrate by parts ∫fg′=fg−∫f′g
f=x g′=ex
f′=1 g=ex
=xex−∫exdx
Now solving ∫exdx
Apply exponential rule,
∫aex=ln(a)ax With a=e
=ex
Plug in solved integrals
=xex−∫exdx=xex−ex
Plug in solved integrals
4∫xexdx=4xex−4ex
The solved problem
∫4xexdx =4xex−4ex+c
=4(x−1)ex+c
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