If $\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+B\log (9e^{2x}-4)+c$ then $36\left(A+B\right)=$

Solution:

$\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+B\log (9e^{2x}-4)+c$

Differentiate w.r.t x on both sides

$\Rightarrow \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}=A+B\frac{9e^{2x}.2}{9e^{2x}-4}$

$\Rightarrow \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}=\frac{9A e^{2x}-4A+18B e^{2x}}{9.e^{2x}-4}$

$\Rightarrow \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}=\frac{9A e^x-4A{e}^{-x}+18B e^x}{9.e^x-4e^{-x}}$

$\Rightarrow$ equate $e^{-x} coefficients on both sides then 6=-4A$

$\Rightarrow A=-3/2$

 $equate e^{x }coefficients on both sides then 4=9A+18B$

$\Rightarrow 4=9\left(-\frac{3}{2}\right)+18B$

$\Rightarrow 4=-\frac{27}{2}+18B$

$\Rightarrow B=\frac{35}{36}$

$\therefore 36\left(A+B\right)=36\left[-\frac{3}{2}+\frac{35}{36}\right]=-19$