\(\displaystyle\int\sqrt{1+\frac{x^2}{9}}\,dx\) is equal to
(Where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{3}\left[\frac{9}{2}\log|x + \sqrt{x^2+9}|+\frac{x}{2}\sqrt{x^2+9}\right]+C\)
Since \(1 + \frac{x^2}{9} = \frac{x^2 + 9}{9}\) and \(\sqrt{9} = 3\):
\( \displaystyle \int\sqrt{1+\frac{x^2}{9}}\,dx = \frac{1}{3}\int\sqrt{x^2 + 9}\,dx \)
By the standard integral
\( \displaystyle \int\sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2+a^2}\right| + C \)
with \(a = 3\):
\( \displaystyle \int\sqrt{1+\frac{x^2}{9}}\,dx = \frac{1}{3}\left[\frac{9}{2}\log\left|x + \sqrt{x^2+9}\right| + \frac{x}{2}\sqrt{x^2+9}\right] + C \)
Tip: The 9 leaves the square root as \(\sqrt{9} = 3\), so the factor in front is \(\frac{1}{3}\), not \(\frac{1}{9}\).
NCERT (2026-27 edition):
Integrals of Some More Types
