If ∫ 0 1 1 5 + 2 x - 2 x 2 1 + e 2 - 4 x d x = 1 α log e α + 1 β , α , β > 0 , then α 4 - β 4 is equal to

Step-by-step solution

Given, ∫ 0 1 1 5 + 2 x - 2 x 2 1 + e 2 - 4 x d x = 1 α log e α + 1 β Now let, I = 1 2 ∫ 0 1 1 5 2 - x 2 - x 1 + e - 4 x - 1 2 d x ⇒ I = 1 2 ∫ 0 1 1 11 4 - x - 1 2 2 1 + e - 4 x - 1 2 d x Now let x - 1 2 = t ⇒ d x = d t So, integral becomes, I = 1 2 ∫ - 1 2 1 2 1 11 2 2 - t 2 1 + e - 4 t d t ⇒ I = 1 2 ∫ - 1 2 0 1 11 2 2 - t 2 1 + e - 4 t d t + 1 2 ∫ 0 1 2 1 11 2 2 - t 2 1 + e - 4 t d t ⏟ I 2 Now solving I 2 = 1 2 ∫ - 1 2 1 2 1 11 2 2 -