\(\int\displaystyle\frac{dx}{\sqrt{49x^2 + 16}}\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{7}\log\left|7x + \sqrt{49x^2 + 16}\right| + C\)
The standard result is
\( \displaystyle \int \frac{dt}{\sqrt{t^2 + a^2}} = \log\left|t + \sqrt{t^2 + a^2}\right| + C \)
Let \(t = 7x\). Then \(dt = 7\,dx\), so \(dx = \frac{1}{7}\,dt\), and \(49x^2 + 16 = t^2 + 4^2\).
\( \displaystyle \int \frac{dx}{\sqrt{49x^2 + 16}} = \frac{1}{7}\int \frac{dt}{\sqrt{t^2 + 16}} \)
\( \displaystyle = \frac{1}{7}\log\left|7x + \sqrt{49x^2 + 16}\right| + C \)
The factor in front is \(\frac{1}{7}\), from the coefficient of x, not \(\frac{1}{4}\).
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{28}\tan^{-1}\left(\frac{7x}{4}\right) + C\) — Incorrect. This is the integral of \(\frac{1}{49x^2 + 16}\), without the square root.
Option (2) \(\log\left|7x + \sqrt{49x^2 + 16}\right| + C\) — Incorrect. The factor \(\frac{1}{7}\) from \(dx = \frac{1}{7}\,dt\) is missing; its derivative is \(\frac{7}{\sqrt{49x^2 + 16}}\).
Option (3) \(\displaystyle\frac{1}{4}\log\left|7x + \sqrt{49x^2 + 16}\right| + C\) — Incorrect. Its derivative is \(\frac{7}{4\sqrt{49x^2 + 16}}\); the factor depends on the 7, not on the 4.