Question

If ∫[(4e^x + 6e^–x)/(9e^x – 4e^–x)]dx = Ax + B log(9e^2x – 4) + c then A, B = _______. 

(a) (3/2), [(– 35)/(36)] 

(b) – (3/2), [(– 35)/(36)] 

(c) – (3/2), (35/36) 

(d) (3/2), (35/36)

Answer 1

The correct option (c) – (3/2), (35/36)   

Explanation:

I = ∫[(4e^x + 6e^–x)/(9e^x + 4e^–x)]dx 

= ∫[(4e^2x + 6)/(9e^2x + 4)]dx 

Let 4e^2x + 6 = A(9e^2x – 4) + B(18e^2x) 

∴ 9A + 18B = 4 – 4A = 6 

∴ A = – (3/2) and B = (35/36) 

∴ ∫[{A(9e^2x – 4) – B(18e^2x)}/(9e^2x – 4)]dx 

= A∫dx + B ∫[(18e^2x)/(9e^2x – 4)]dx 

= – (3/2)x + (35/36)log|9e^2x – 4| + c 

∴ A = – (3/2) and B = (35/36)