Question

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

• A. $-\frac{19}{36}$
• B. $\frac{19}{36}$
• C. $\frac{36}{19}$
• D. $-\frac{36}{19}$

Answer 1

The correct option is A $-\frac{19}{36}$

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

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

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

Now differentiate both sides, we get

$\frac{4e^{2x}+6}{9e^{2x}-4}=A+\frac{B\left(18e^{2x}\right)}{9e^{2x}-4}$

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

$⟹4e^{2x}+6=(9A+18B)e^{2x}-4A$

On comparing, we get
$A=-\frac{3}{2},B=\frac{35}{36},C\in R$

$\therefore A+B=-\frac{3}{2}+\frac{35}{36}=-\frac{19}{36}$