Question

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

• A. -3/2
• B. 2/3
• C. -1/3
• D. -2/3

Answer 1

The correct option is A -3/2

$\int \frac{4e^{2x}+6}{9e^{2x}-4}\cdot dx$
Suppose $N\left(x\right)=4e^x+6e^{-x}=4e^{2x}+6$,
and $D\left(x\right)=9e^x-4e^{-x}=9e^{2x}+4$.
$4e^{2x}+6=A\left(D^1\right(x\left)\right)+B\left(D\right(x\left)\right)$
$4e^{2x}+6=A\left(18e^{2x}\right)+B(9e^{2x}-4)$
$-4B=6$
$B=\frac{-3}{2};A=\frac{4}{27}$
$N\left(x\right)=\frac{4}{27}\left(D^1\right(x\left)\right)-\frac{3}{2}\left(D\right(x\left)\right)$
$\int \frac{N\left(x\right)}{D\left(x\right)}dx=\int \frac{\frac{4}{27}{D}^1\left(x\right)}{D\left(x\right)}-\frac{3}{2}\int \frac{D\left(x\right)}{D\left(x\right)}dx+c$
$=\frac{4}{27}log\left[D\right(x\left)\right]-\frac{3}{2}x+c.$
$=\frac{4}{27}log[9e^{2x}-4]-\frac{3}{2}x+c$
$A=\frac{-3}{2}$