\(\int\displaystyle\frac{\sqrt{16 + (\log x)^2}}{x}\,dx\) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(8\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + \displaystyle\frac{\log x}{2}\sqrt{16 + (\log x)^2} + C\)
Let
\(\displaystyle t = \log x,\quad dt = \frac{dx}{x}\)
\(\displaystyle \int\frac{\sqrt{16 + (\log x)^2}}{x}\,dx = \int\sqrt{4^2 + t^2}\,dt\)
By the standard integral
\(\displaystyle \int\sqrt{a^2 + t^2}\,dt = \frac{t}{2}\sqrt{a^2 + t^2} + \frac{a^2}{2}\log\left|t + \sqrt{a^2 + t^2}\right| + C\)
with a = 4, so that \(\frac{a^2}{2} = 8\):
\(\displaystyle \int\sqrt{16 + t^2}\,dt = \frac{t}{2}\sqrt{16 + t^2} + 8\log\left|t + \sqrt{16 + t^2}\right| + C\)
Putting back \(t = \log x\):
\(\displaystyle = 8\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + \frac{\log x}{2}\sqrt{16 + (\log x)^2} + C\)
Why the other options are incorrect
Option (1) \(2\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + \log x\sqrt{16 + (\log x)^2} + C\) — Incorrect. Its coefficients are 2 and 1; the formula gives 8 and \(\frac{1}{2}\).
Option (2) \(16\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + \frac{\log x}{2}\sqrt{16 + (\log x)^2} + C\) — Incorrect. The log term has \(a^2 = 16\) without the factor \(\frac{1}{2}\).
Option (4) \(4\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + \frac{\log x}{2}\sqrt{16 + (\log x)^2} + C\) — Incorrect. The log term has a = 4 instead of \(\frac{a^2}{2} = 8\).