Question

If ∫[(2e^x + 3e^–x)/(3e^x + 4e^–x)]dx = Ax + Blog|3e^2x + 4| + c then A + B = ________

(a) (11/24)

(b) (13/24)

(c) (15/24)

(d) (17/24)

Answer 1

The correct option (d) (17/24)   

Explanation:

[(2e^x + 3e^–x)/(3e^x + 4e^–x)] = [(2e^2x + 3)/(3e^2x + 4)]

∴   2e^2x + 3 = A(3e^2x + 4) + B(6 ∙ e^2x)

∴  2 = 3A + 6B   &   3 = 4A ⇒ A = (3/4)

∴   6B = 2 – 3A = 2 – (9/4) = [(– 1)/4] ⇒ B = – (1/24)

∴   I = ∫[(2e^2x + 3)/(3e^2x + 4)]dx

= ∫[({+ (3/4)(3e^2x + 4) – (1/24)(6 ∙ e^2x)dx})/(3e^2x + 4)]dx

= (3/4)∫dx – (1/24)∫[(6 ∙ e^2x)/(3e^2x + 4)]dx

= (3/4)x – (1/24) log|3e^2x + 4| + c

by comparing,

A = (3/4) and B = – (1/24)

∴  A + B = (17/24)