Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\displaystyle\frac{e^{2x}}{1+e^{2x}}\,dx\) | (I) | \(\tan^{-1}\left(e^x\right)+C\) |
| (B) | \(\int\displaystyle\frac{e^x}{1+e^{2x}}\,dx\) | (II) | \(\displaystyle -\frac{1}{1+e^x}+C\) |
| (C) | \(\int\displaystyle\frac{e^x}{\left(1+e^x\right)^2}\,dx\) | (III) | \(\displaystyle \frac{1}{2}\log\left(1+e^{2x}\right)+C\) |
| (D) | \(\int e^x\left(1+e^x\right)^2dx\) | (IV) | \(\displaystyle \frac{1}{3}\left(1+e^x\right)^3+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
All four look alike, but the right substitution depends on where \(e^x\) and \(e^{2x}\) stand.
(A) \(\int\frac{e^{2x}}{1+e^{2x}}\,dx\)
This matches (III) \(\frac{1}{2}\log\left(1+e^{2x}\right)+C\).
The numerator is half the derivative \(2e^{2x}\) of the denominator (\(\frac{f'}{f}\) form):
\( \displaystyle \frac{1}{2}\int\frac{2e^{2x}}{1+e^{2x}}\,dx = \frac{1}{2}\log\left(1+e^{2x}\right) \)
(B) \(\int\frac{e^x}{1+e^{2x}}\,dx\)
This matches (I) \(\tan^{-1}\left(e^x\right)+C\).
\( \displaystyle t = e^x:\quad \int\frac{dt}{1+t^2} = \tan^{-1}t \)
(C) \(\int\frac{e^x}{\left(1+e^x\right)^2}\,dx\)
This matches (II) \(-\frac{1}{1+e^x}+C\).
\( \displaystyle t = 1+e^x:\quad \int\frac{dt}{t^2} = -\frac{1}{t} \)
(D) \(\int e^x\left(1+e^x\right)^2dx\)
This matches (IV) \(\frac{1}{3}\left(1+e^x\right)^3+C\).
\( \displaystyle t = 1+e^x:\quad \int t^2\,dt = \frac{t^3}{3} \)