Let f = integral symbol.
f((x- 1/2)e^-2x)dx
Break it up
Let u = x-1/2
let du=x-1/2 dx
let dw =e^-2x
let w= -2e^2x
So basically fu dw= uw-fwdu (got it on paper but difficult to
transfer)
=f((x- 1/2)e^-2x)dx
= (x-1/2)(-2e^2x)- f(-2e^2x)(x-1/2 dx)
= (few more steps same as the uw-fwdu)
= -1/2(e^(2x))x
Law of exponential integrals
fe^x dx = e^x +C
Not sure if I got right then again ...I got no A levels....so you
have to give way .
