Question

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

• A. $(\frac{-19}{36},\frac{35}{36})$
• B. $(\frac{-3}{2},\frac{35}{36})$
• C. $(\frac{19}{36},\frac{-35}{36})$
• D. $(\frac{3}{2},\frac{-35}{36})$

Answer 1

The correct option is B $(\frac{-3}{2},\frac{35}{36})$

$\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=\int \frac{2(2e^{2x}+3)}{(9e^{2x}-4)}dx$

Let, $e^{2x}=t⟹2e^{2x}dx=dt⟹2dx=\frac{dt}{t}$

$=\int \frac{2t+3}{t(9t-4)}dt$

$\because \frac{2t+3}{t(9t-4)}=\frac{A}{t}+\frac{B}{9t-4}$

$⟹2t+3=A(9t-4)+Bt$

$⟹2t+3=(9A+B)t-4A$

$⟹-4A=3$ and $9A+B=2$

$⟹A=\frac{-3}{4}$ and $\frac{-27}{4}+B=2$

$⟹A=\frac{-3}{4}$ and $B=2+\frac{27}{4}=\frac{35}{4}$

$=\int (\frac{-3}{4t}+\frac{35}{4(9t-4)})dt$

$=\frac{-3}{4}\int \frac{1}{t}dt+\frac{35}{4}\int \frac{1}{9t-4}dt$

$=\frac{-3}{4}{\log }_{e}t+\frac{35}{4}\frac{1}{9}{\log }_e|9t-4|+c$

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

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

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

$\therefore \int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx$$=\frac{-3x}{2}+\frac{35}{36}\log |9e^{2x}-4|+c$

$(A,B)=(\frac{-3}{2},\frac{35}{36})$