\(\int\displaystyle\frac{3x-1}{\sqrt{x^2+9}}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3\sqrt{x^2+9}-\log\left|x+\sqrt{x^2+9}\right|+C\)
The integral splits into two parts:
\( \displaystyle I = 3\int\frac{x}{\sqrt{x^2+9}}\,dx-\int\frac{dx}{\sqrt{x^2+9}} \)
For the first part, with \(t = x^2+9\), \(dt = 2x\,dx\):
\( \displaystyle 3\int\frac{x}{\sqrt{x^2+9}}\,dx = \frac{3}{2}\int\frac{dt}{\sqrt{t}} = 3\sqrt{x^2+9} \)
The second part is the standard integral \(\int\frac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\) with a = 3:
\( \displaystyle \int\frac{dx}{\sqrt{x^2+9}} = \log\left|x+\sqrt{x^2+9}\right| \)
So
\( I = 3\sqrt{x^2+9}-\log\left|x+\sqrt{x^2+9}\right|+C \)
Why the other options are incorrect
Option (1) \(3\sqrt{x^2+9}+\log\left|x+\sqrt{x^2+9}\right|+C\) — Incorrect. The constant in the numerator is \(-1\), so the log term is subtracted.
Option (3) \(\frac{3}{2}\sqrt{x^2+9}-\log\left|x+\sqrt{x^2+9}\right|+C\) — Incorrect. It forgets that \(\int\frac{dt}{\sqrt{t}} = 2\sqrt{t}\), which doubles \(\frac{3}{2}\) to 3.
Option (4) \(3\sqrt{x^2+9}-\tan^{-1}\left(\frac{x}{3}\right)+C\) — Incorrect. \(\tan^{-1}\) comes from \(\frac{1}{x^2+9}\) without a square root; here the root is present.