how do you integrate 2e^2x/(4-x)

Question

integration

Answer 1

I=∫(2e²ˣ/(4-x))dx.

Let u=4-x, du=-dx, x=4-u.

I=-2∫(e²⁽⁴⁻ᵘ⁾/u)du.

e²⁽⁴⁻ᵘ⁾=e⁸e⁻²ᵘ so I=-2e⁸∫(e⁻²ᵘ/u)du.

e⁻²ᵘ=1-2u+4u²/2!-8u³/3!+...

e⁻²ᵘ/u=1/u-2+4u/2!-8u²/3!+...

∫(e⁻²ᵘ/u)du=ln|u|-2u+4u²/(2(2!))-8u³/(3(3!))+...

I=C-2e⁸[ln|x-4|+∑2ʳ(x-4)ʳ/(r(r!))] for r∊[1,∞)  where C is a constant of integration.