Question

If \(a > 0,b > 0\) and \(\int \frac{1}{ax^2+b}dx = \frac{1}{\sqrt{6}}tan^{-1}(\frac{\sqrt{2}x}{\sqrt{3}})+c\), then \(\int \frac{1}{bx^2+a}dx = \ldots\)

• A. \(-\frac{1}{\sqrt{6}}tan^{-1}(\frac{\sqrt{2}x}{\sqrt{3}})+c\)
• B. \(\frac{1}{\sqrt{6}}tan^{-1}(\frac{\sqrt{3}x}{\sqrt{2}})+c\)
• C. \(-\sqrt{6}\,tan^{-1}(\frac{\sqrt{2}x}{\sqrt{3}})+c\)
• D. \(\sqrt{6}\,tan^{-1}(\frac{\sqrt{3}x}{\sqrt{2}})+c\)

Hint:

Match the standard form to find a and b, then swap them.

Answer 1

The Correct Option is B

Step 1: Find a and b
The coefficient $\frac{1}{\sqrt{ab}} = \frac1{\sqrt6}$ and the argument factor $\sqrt{a/b} = \sqrt{2/3}$. Hence $a=2, b=3$.

Step 2: Swap
For $\frac{1}{bx^2+a}$ the roles swap: factor $\sqrt{b/a} = \sqrt{3/2}$ and coefficient is still $\frac{1}{\sqrt6}$.

Step 3: Result
Option (B).

Final Answer:
Option B. \[ \boxed{\text{(B)}\ \frac1{\sqrt6}\tan^{-1}\left(\frac{\sqrt3x}{\sqrt2}\right)+c} \]