The value of \(\int_{0}^{1}\displaystyle\frac{x}{x+1}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1-\log_e 2\)
Writing the numerator as \((x+1) - 1\):
\( \displaystyle \frac{x}{x+1} = \frac{(x+1)-1}{x+1} = 1 - \frac{1}{x+1} \)
So
\( \displaystyle \int_{0}^{1}\left(1 - \frac{1}{x+1}\right)dx = \left[x - \log_e(x+1)\right]_{0}^{1} \)
\( = (1 - \log_e 2) - (0 - 0) \)
\( = 1 - \log_e 2 \)
Why the other options are incorrect
Option (1) \(\log_e 2\) — Incorrect. This is \(\int_{0}^{1}\frac{1}{x+1}\,dx\) alone; the term 1 is left out.
Option (2) \(1+\log_e 2\) — Incorrect. The fraction \(\frac{1}{x+1}\) is subtracted, not added.
Option (4) \(\log_e 2-1\) — Incorrect. This is negative (about −0.31), but \(\frac{x}{x+1} \ge 0\) on [0, 1], so the integral is positive.