We will rewrite:

e^x * (x-2) = x*e^x - 2*e^x

==> int x*e^x dx - 2Int e^x

But we know that:

==> int u*dv = u*v - int v du

==> u= x dv= e^x dx

==> du = dx ==> v = e^x

==> x*e^x - int e^x dx

...

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We will rewrite:

e^x * (x-2) = x*e^x - 2*e^x

==> int x*e^x dx - 2Int e^x

But we know that:

==> int u*dv = u*v - int v du

==> u= x dv= e^x dx

==> du = dx ==> v = e^x

==> x*e^x - int e^x dx

==> x*e^x - e^x = e^x ( x-1) - 2*e^x

==> e^x ( x-1-2) = e^x *(x-3).

**==> Int e^x * (x-2) dx = e^x * (x-3).**

We have to integrate e^x / (x – 2)

We use integration by parts here:

The general formula is Int [ u dv] = u*v – Int [ v du]

let u = (x – 2) , du = dx

dv = e^x , v = e^x

=> Int [ (x – 2) * e^x dx] = (x – 2)* e^x – Int [ e^x dx]

=> Int [ e^x * ( x – 2) dx] = (x – 2) e^x – e^x

=> Int [ e^x * ( x - 2) dx] = e^x ( x – 3) + C

**The integral of e^x / (x-2) is e^x ( x – 3) + C**