Q88 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $a > 0, b > 0$ and $\int\dfrac{1}{ax^2 + b}dx = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$, then $\int\dfrac{1}{bx^2 + a}dx = \ldots$
- A$-\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$
- Correct answerB$\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
- C$-\sqrt{6}\,\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$
- D$\sqrt{6}\,\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
Solution
We know that $\int\dfrac{1}{ax^2 + b}dx = \dfrac{1}{\sqrt{ab}}\tan^{-1}\left(\sqrt{\dfrac{a}{b}}x\right) + c$
Comparing this with the given expression $\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$, we get:
$\dfrac{1}{\sqrt{ab}} = \dfrac{1}{\sqrt{6}} \Rightarrow ab = 6$
$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{2}}{\sqrt{3}} \Rightarrow \dfrac{a}{b} = \dfrac{2}{3}$
Multiplying the two equations, we get $a^2 = 4 \Rightarrow a = 2$ (since $a > 0$).
Substituting $a = 2$ in $ab = 6$, we get $b = 3$.
Now, we need to evaluate $\int\dfrac{1}{bx^2 + a}dx$:
$\int\dfrac{1}{3x^2 + 2}dx = \dfrac{1}{\sqrt{3 \times 2}}\tan^{-1}\left(\sqrt{\dfrac{3}{2}}x\right) + c$
$= \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
Answer: $\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
Comparing this with the given expression $\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$, we get:
$\dfrac{1}{\sqrt{ab}} = \dfrac{1}{\sqrt{6}} \Rightarrow ab = 6$
$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{2}}{\sqrt{3}} \Rightarrow \dfrac{a}{b} = \dfrac{2}{3}$
Multiplying the two equations, we get $a^2 = 4 \Rightarrow a = 2$ (since $a > 0$).
Substituting $a = 2$ in $ab = 6$, we get $b = 3$.
Now, we need to evaluate $\int\dfrac{1}{bx^2 + a}dx$:
$\int\dfrac{1}{3x^2 + 2}dx = \dfrac{1}{\sqrt{3 \times 2}}\tan^{-1}\left(\sqrt{\dfrac{3}{2}}x\right) + c$
$= \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
Answer: $\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
Hence, the correct answer is option: 'B'.