Q109 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics
If $\int\left(\frac{4 e^x+6 e^{-x}}{9 e^x-4 e^{-x}}\right) d x=A x+B \log \left|\left(9 e^{2 x}-4\right)\right|+C$, then $(\mathrm{A}, \mathrm{B})=$
- A$\left(\frac{3}{2}, \frac{35}{36}\right)$
- Correct answerB$\left(\frac{-3}{2}, \frac{-35}{36}\right)$
- C$\left(\frac{-3}{2}, \frac{35}{36}\right)$
- D$\left(\frac{3}{2}, \frac{-35}{36}\right)$
Solution
Let $I=\int \frac{4 e^x+6 e^{-x}}{9 e^x-4 e^{-x}} d x=\int \frac{4 e^{2 x}+6}{9 e^{2 x}-4} d x$ Let $4 e^{2 x}+6=A\left(9 e^{2 x}-4\right)+B\left(18 e^{2 x}\right)$ $4 e^{2 x}+6=(9 A+18 B) e^{2 x}-4 A$ Comparing both sides we get, $A=-\frac{3}{2}, B=\frac{35}{36}$
$\begin{aligned} & \therefore I=-\frac{3}{2} \int \frac{\left(9 e^{2 x}-4\right) d x}{\left(9 e^{2 x}-4\right)}+\frac{35}{36} \int \frac{18 e^{2 x}}{9 e^{2 x}-4} d x \\ & \Rightarrow \quad I=-\frac{3}{2} \int d x+\frac{35}{36} \int \frac{18 e^{2 x}}{9 e^{2 x}-4} d x \\ & \Rightarrow \quad I=-\frac{3}{2}+\frac{35}{36} \log \left|\left(9 e^{2 x}-4\right)\right|+c \\ & \text { Hence } A=-\frac{3}{2}, \quad B=\frac{35}{36}\end{aligned}$
Hence, the correct answer is option: 'B'.