Question

If I = (4e^x + 6e^–x /9e^x – 4e^–x) dx = Ax + B log |(9e^2x – 4)| + C then A = ..., B = ... and C = ...

Answer 1

4e^x + 6e^–x = m(9e^x – 4e^–x) + n(9e^x + 4e^–x) 

Comparing the co-efficients of e^x and e^–x, we get 

m + n = 4/9 and n – m = 3/2 

Solving, we get, 

n = 35/36 and m = –19/36. 

The given integral reduces to