Q110 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
$\int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x=K x+L \log \left(e^{-2 x}+\frac{7}{3}\right)+C$
then $K+L=$
- A$\frac{-3}{38}$
- B$\frac{21}{38}$
- Correct answerC$\frac{38}{21}$
- D$\frac{-38}{3}$
Solution
$\begin{aligned} I & =\int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x \\ & =3 \int \frac{e^x-\frac{7}{3} e^{-x}}{7 e^x+3 e^{-x}} d x \\ & =\frac{3}{7} \int \frac{7 e^x-\frac{49}{3} e^{-x}}{7 e^x+3 e^{-x}} d x \\ & =\frac{3}{7} \int \frac{\left(7 e^x+3 e^{-x}\right)-\frac{58}{3} e^{-x}}{7 e^x+3 e^{-x}} d x \\ & =\frac{3}{7} \int d x-\frac{3}{7} \times \frac{58}{3} \int \frac{e^{-x}}{7 e^x+3 e^{-x}} d x\end{aligned}$
$I=\frac{3 x}{7}-\frac{58}{7} \int \frac{e^{-2 x}}{7+3 e^{-2 x}} d x$
Put $t=7+3 e^{-2 x}$
$\Rightarrow \quad d t=-6 e^{-2 x} d x$
$\begin{aligned} & I=\frac{3 x}{7}-\frac{58}{7}\left(-\frac{1}{6}\right) \int \frac{d t}{t} \\ & I=\frac{3 x}{7}+\frac{29}{21} \ln |t|+C \\ & I=\frac{3 x}{7}+\frac{29}{21} \ln \left(7+3 e^{-2 x}\right)+C\end{aligned}$
$\begin{aligned} & \Rightarrow \quad K=3 / 7 \text { and } L=29 / 21 \\ & \Rightarrow K+L=\frac{3}{7}+\frac{29}{21}=\frac{9+29}{21}=\frac{38}{21}\end{aligned}$
Hence, the correct answer is option: 'C'.