MHT CET · Maths · Indefinite Integration
If \(\int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=A x+B \log \left(3 e^{2 x}+4\right)+C\), then values of \(A\) and \(B\) are respectively (where \(C\) is a constant of integration.)
- A \(\frac{3}{4}, \frac{-1}{24}\)
- B \(\frac{3}{4}, \frac{1}{24}\)
- C \(\frac{4}{3},-24\)
- D \(\frac{1}{4}, \frac{1}{24}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{4}, \frac{-1}{24}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=\int \frac{2 e^{2 x}+3}{3 e^{2 x}+4} d x \\ & \int \frac{\frac{3}{4}\left(3 e^{2 x}+4\right)-\frac{1}{24}\left(6 e^{2 x}\right)}{3 e^{2 x}+4} d x \\ & =\frac{3}{4} x-\frac{1}{24} \log \left(3 e^{2 x}+4\right)+c \\ & \Rightarrow A=\frac{3}{4} \text { and } B=\frac{-1}{24}\end{aligned}\)
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