TS EAMCET · Maths · Indefinite Integration
If \(\int \frac{e^x-1}{e^x+1} d x=f(x)+c\), then \(f(x)\) is equal to
- A \(2 \log \left(e^x+1\right)\)
- B \(\log \left(e^{2 x}-1\right)\)
- C \(2 \log \left(e^x+1\right)-x\)
- D \(\log \left(e^{2 x}+1\right)\)
Answer & Solution
Correct Answer
(C) \(2 \log \left(e^x+1\right)-x\)
Step-by-step Solution
Detailed explanation
We have, \(\int \frac{e^x-1}{e^x+1} d x\) \(=\int\left(\frac{2 e^x}{e^x+1}-1\right) d x\) \(=2 \log \left(e^x+1\right)-x+c\) On comparing with \(f(x)+c\), we get \(f(x)=2 \log \left(e^x+1\right)-x\)
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