Question

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

• A. $A=\frac{3}{2}$
• B. $B=\frac{35}{36}$
• C. C is indefinite
• D. $A+B=\frac{-19}{36}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $B=\frac{35}{36}$
C $A+B=\frac{-19}{36}$
D C is indefinite

Let $I=\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=\int \frac{4e^{2x}+6}{9e^{2x}-4}dx$

Let $4e^{2x}+6=A(9e^{2x}-4)+B\left(18e^{2x}\right)$
$\Rightarrow 9A+18B=4$ and $-4A=6$
$\Rightarrow A=-\frac{3}{2}$ and $B=\frac{35}{36}$

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

$=A\int dx+B\int \frac{1}{t}dt$ where $t=9e^{2x}-4$

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

$=-\frac{3}{2}+\frac{35}{36}\log (9e^{2x}-4)+c$

Gives $A=-\frac{3}{2},B=\frac{35}{36}$ and $C=$ any real number

$A+B=\frac{-19}{36}$