The value of \(\int_{0}^{1}\displaystyle\frac{dx}{4-x^{2}}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{4}\log_e 3\)
Using the standard result
\( \displaystyle \int\frac{dx}{a^{2}-x^{2}} = \frac{1}{2a}\log_e\left|\frac{a+x}{a-x}\right| + C \)
with \(a = 2\):
\( \displaystyle \int_{0}^{1}\frac{dx}{4-x^{2}} = \frac{1}{4}\left[\log_e\frac{2+x}{2-x}\right]_{0}^{1} \)
\( \displaystyle = \frac{1}{4}\left(\log_e\frac{3}{1} - \log_e\frac{2}{2}\right) \)
\( \displaystyle = \frac{1}{4}\left(\log_e 3 - 0\right) = \frac{1}{4}\log_e 3 \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}\log_e 3\) — Incorrect. The factor is \(\frac{1}{2a} = \frac{1}{4}\) for \(a = 2\), not \(\frac{1}{2}\).
Option (2) \(\displaystyle\frac{1}{4}\log_e\frac{1}{3}\) — Incorrect. This is negative, but \(\frac{1}{4-x^{2}} \gt 0\) on \([0, 1]\); the fraction inside the log has been turned upside down.
Option (3) \(\log_e 3\) — Incorrect. It drops the factor \(\frac{1}{2a} = \frac{1}{4}\).