Question

If $\int \limits \frac {4e^x+6e^{-x}}{9e^x-4e^-x}dx= Ax+ B log (9e^{2x}-4)+ C,$ then A = ..., B=... and C=... .

Answer 1

Given, $\int \limits \frac {4e^x+6e^{-x}}{9e^x-4e^{-x}}dx = Ax+ B \, log (9e^{2x}-4) + c $
$ LHS = \int \limits \frac {4e^{2x}+6}{9e^{2x}-4}dx$
Let $ 4e^{2x}+6 = A (9e^{2x}-4)+ B (18e^{2x})$
$\Rightarrow 9A+18B=4 \, \, and \, \, -4A = 6$
$\Rightarrow A=- \frac {3}{2} \, \, \, \, and \, \, \, \, B= \frac {35}{36}$
$\therefore \int \limits \frac {A(9e^{2x}-4)+B(18e^{2x})}{9e^{2x}-4}dx$
$ =A \int \limits 1dx+B \int \limits \frac {1}{t}dt , \, where \, t =9e^{2x}-4$
$ =Ax+B \, log (9e^{2x}-4)+c$
$ = - \frac {3}{2}x+ \frac {35}{36} log (9e^{2x}-4)+c$
$\therefore A=- \frac {3}{2}, B= \frac {35}{36}$
and c=any real number