Question

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

Answer 1

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