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: Standard form
\(\int\frac{dx}{ax^2+b} = \frac{1}{\sqrt{ab}}\tan^{-1}\left(x\sqrt{\frac ab}\right)+c\).

Step 2: Compare
So \(\sqrt{ab} = \sqrt6\), giving \(ab = 6\), and \(\frac ab = \frac23\). Multiplying, \(a^2 = 4\), so \(a = 2\) and \(b = 3\).

Step 3: Required integral
\(\int\frac{dx}{3x^2+2} = \frac{1}{\sqrt6}\tan^{-1}\left(x\sqrt{\frac32}\right)+c = \frac1{\sqrt6}\tan^{-1}\left(\frac{\sqrt3x}{\sqrt2}\right)+c\).

Step 4: Result
Option (B). The sign cannot be negative since the integrand is positive, which removes (A) and (C).

Final Answer:
The integral is (1/root 6) tan^-1 (root3 x / root2) + c. \[ \boxed{\text{(B)}\ \frac1{\sqrt6}\tan^{-1}\left(\frac{\sqrt3x}{\sqrt2}\right)+c} \]