The value of \(\int_{0}^{1}\displaystyle\frac{dx}{2x+1}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\log_e 3\)
Since \(\displaystyle\int\frac{dx}{ax+b} = \frac{1}{a}\log_e|ax+b| + C\), with \(a = 2\):
\( \displaystyle \int_{0}^{1}\frac{dx}{2x+1} = \frac{1}{2}\left[\log_e(2x+1)\right]_{0}^{1} \)
\( \displaystyle = \frac{1}{2}\left(\log_e 3 - \log_e 1\right) \)
\( \displaystyle = \frac{1}{2}\log_e 3 \)
Why the other options are incorrect
Option (1) \(\log_e 3\) — Incorrect. It leaves out the factor \(\frac{1}{2}\) that comes from the coefficient 2 of x.
Option (2) \(2\log_e 3\) — Incorrect. It multiplies by 2 instead of dividing by 2.
Option (3) \(\log_e 2\) — Incorrect. At \(x = 1\), \(2x+1 = 3\), not 2.