Integration M2

(c) Let f(x) = 1/(x^2 - 3x + 3)(e^(2x-3) + 1)
f(3 - x) = 1/[(3-x)^2 - 3(3-x) + 3][e^(3-2x) + 1]
= 1/[(x^2 - 3x + 3)(e^(3-2x) + 1)]
= e^(2x-3)/[(x^2 - 3x + 3)(e^(2x-3) + 1)]

f(x) + f(3 - x) = [1 + e^(2x-3)]/[(x^2 - 3x + 3)(e^(2x-3) + 1)]
= 1/(x^2 - 3x + 3)
Hence ∫f(x) + f(3 - x) dx [0,3] = (4√3)π/9 [by (b)]
Since ∫f(x) dx [0,3] = 1/2 ∫[f(x) + f(3-x)]dx [0,3] [by (a)]
∫dx/(x^2 - 3x + 3)(e^(2x-3) + 1) [0,3] = ((4√3)π)/9) / 2 = (2√3)π/9