MHT CET · Maths · Indefinite Integration
\(\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=\)
- A \(x-\log \left(1-2 e^{-x}\right)+\mathrm{c}\)
- B \(\log \left(1-2 e^{-x}\right)+\mathrm{c}\)
- C \(x+\log \left(1-2 e^{-x}\right)+\mathrm{c}\)
- D \(x+2\log \left(1-2 e^{-x}\right)+\mathrm{c}\)
Answer & Solution
Correct Answer
(D) \(x+2\log \left(1-2 e^{-x}\right)+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
(B)
Let \(1 =\int \frac{1-2 e^{-x}}{1-2 e^{-1}} \)
\(=\int \frac{1-2 e^{-x}+4 e^{-x}}{1-2 e^{-1}} d x=\int \frac{1-2 e^{-x}}{1-2 e^{-1}} d x+4 \int \frac{e^{-x}}{1-2 e^{-x}} d x \)
\(=\int d x+\frac{4}{2} \int \frac{2 e^{-x}}{1-2 e^{-x}} d x=x-2 \cdot \log 1-2 e^{-1}-c\)
Let \(1 =\int \frac{1-2 e^{-x}}{1-2 e^{-1}} \)
\(=\int \frac{1-2 e^{-x}+4 e^{-x}}{1-2 e^{-1}} d x=\int \frac{1-2 e^{-x}}{1-2 e^{-1}} d x+4 \int \frac{e^{-x}}{1-2 e^{-x}} d x \)
\(=\int d x+\frac{4}{2} \int \frac{2 e^{-x}}{1-2 e^{-x}} d x=x-2 \cdot \log 1-2 e^{-1}-c\)
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