If a > 0, b > 0 and {1}{a x²+b} d x={1}{{6}} ^{-1}≤ft({{2} x}{{3}})+c, then {1}{b x²+a} d x=

Correct Answer is : 16tan⁡−1(3x2)+c{1}{{6}} ^{-1}≤ft({{3} x}{{2}})+c61tan−1(23x)+c

-{1}{{6}} ^{-1}≤ft({{2} x}{{3}})+c

{1}{{6}} ^{-1}≤ft({{3} x}{{2}})+c

-{6} ^{-1}≤ft({{2} x}{{3}})+c

Correct Answer is : 16tan⁡−1(3x2)+c{1}{{6}} ^{-1}≤ft({{3} x}{{2}})+c61tan−1(23x)+c

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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 129 of 150

Marks: +1, -0

a > 0, b > 0

\int \frac{1}{a x^{2}+b} d x=\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{\sqrt{2} x}{\sqrt{3}}\right)+c

\int \frac{1}{b x^{2}+a} d x=\ldots

Solution:

Concept:

Use the standard formula:

\int \frac{1}{a x^{2}+b}\, dx = \frac{1}{\sqrt{ab}} \tan^{-1}\left(x\sqrt{\frac{a}{b}}\right) + C

This formula is valid for

a > 0

b > 0

Explanation:

Given:

\int \frac{1}{a x^{2}+b}\, dx = \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{3}}\right) + C

Compare with the standard formula:

\frac{1}{\sqrt{ab}} = \frac{1}{\sqrt{6}} \Rightarrow ab = 6

Also,

x\sqrt{\frac{a}{b}} = \frac{\sqrt{2}x}{\sqrt{3}} \Rightarrow \frac{a}{b} = \frac{2}{3}

Solving

ab = 6

and

\frac{a}{b} = \frac{2}{3}

, we get:

a = 2, \quad b = 3

Now,

\int \frac{1}{b x^{2}+a}\, dx = \int \frac{1}{3x^{2}+2}\, dx

Using the same formula with

a = 3

b = 2

\int \frac{1}{3x^{2}+2}\, dx = \frac{1}{\sqrt{6}} \tan^{-1}\left(x\sqrt{\frac{3}{2}}\right) + C

Therefore,

\int \frac{1}{b x^{2}+a}\, dx = \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{\sqrt{3}x}{\sqrt{2}}\right) + C

Answer:

Option B:

\frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{\sqrt{3}x}{\sqrt{2}}\right) + C

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