CUET · MATHS · PYQ PAPER 2023
The integral \(I=\int \frac{e^x}{1-e^{2 x}} d x\) is equal to :
- A \(\frac{1}{2}\left[-\log \left(1-e^x\right)+\log \left(1+e^x\right)\right]+C\)
- B \(\log \left(1-2 e^x\right)+\log \left(1+2 e^x\right)+C\)
- C \(\frac{1}{2}\left[\log \left(x-e^x\right)+\log \left(x+e^x\right)\right]+C\)
- D \(\log \left(1-x-e^{-x}+e^x\right)+C\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\left[-\log \left(1-e^x\right)+\log \left(1+e^x\right)\right]+C\)
Step-by-step Solution
Detailed explanation
Let \(u = e^x\), then \(du = e^x dx\). \(I = \int \frac{du}{1-u^2} = \frac{1}{2} \log \left( \frac{1+u}{1-u} \right) + C\) \(I = \frac{1}{2} \log \left( \frac{1+e^x}{1-e^x} \right) + C\) \(I = \frac{1}{2}\left[\log \left(1+e^x\right) - \log \left(1-e^x\right)\right]+C\)…
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