Correct Answer is : 13tan⁡−1(4x2−93)+c1/3 ^{-1}≤ft({{4 x²-9}}{3})+c31tan−1(34x2−9)+c

4/3 ^{-1}≤ft({{4 x²-9}}{3})+c

1/3 ^{-1}≤ft({{4 x²-9}}{3})+c

Correct Answer is : 13tan⁡−1(4x2−93)+c1/3 ^{-1}≤ft({{4 x²-9}}{3})+c31tan−1(34x2−9)+c

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COMEDK 2025 Afternoon Solved Paper

Question : 169 of 180

Marks: +1, -0

\int \frac{d x}{x \sqrt{4 x^2-9}}=

Solution:

Concept:

Use trigonometric substitution to simplify the integrand involving

\sqrt{4x^2-9}

Explanation:

Substitute

x = \frac{3}{2}\sec\theta

Then

dx = \frac{3}{2}\sec\theta\tan\theta\,d\theta

Also

\sqrt{4x^2-9} = \sqrt{9\sec^2\theta - 9} = 3\tan\theta

The integral becomes:

\int \frac{dx}{x\sqrt{4x^2-9}} = \int \frac{\frac{3}{2}\sec\theta\tan\theta\,d\theta}{\left(\frac{3}{2}\sec\theta\right)(3\tan\theta)} = \int \frac{1}{3}\,d\theta = \frac{\theta}{3} + C.

Since

\tan\theta = \frac{\sqrt{4x^2-9}}{3}

, we have

\theta = \tan^{-1}\left(\frac{\sqrt{4x^2-9}}{3}\right)

Thus the integral simplifies to

\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{4x^2-9}}{3}\right) + C

Answer:

Option D:

\frac{1}{3} \tan ^{-1}\left(\frac{\sqrt{4 x^2-9}}{3}\right)+c

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