Q63 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics

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

  • A
    $\dfrac{2}{3} \log \left|\dfrac{x+3}{x-3}\right|+c$
  • B
    $\dfrac{4}{3} \tan ^{-1}\left(\dfrac{\sqrt{4 x^2-9}}{3}\right)+c$
  • Correct answer
    C
    $\dfrac{1}{3} \tan ^{-1}\left(\dfrac{\sqrt{4 x^2-9}}{3}\right)+c$
  • D
    $\dfrac{2}{3} \log \left|\dfrac{x-3}{x+3}\right|+c$

Solution

Let $I = \int \dfrac{dx}{x \sqrt{4x^2 - 9}}$.

Substitute $x = \dfrac{3}{2} \sec \theta$, then $dx = \dfrac{3}{2} \sec \theta \tan \theta \, d\theta$.

The expression $\sqrt{4x^2 - 9}$ becomes $\sqrt{4(\dfrac{9}{4} \sec^2 \theta) - 9} = \sqrt{9(\sec^2 \theta - 1)} = 3 \tan \theta$.

Substituting these into the integral:

$I = \int \dfrac{\dfrac{3}{2} \sec \theta \tan \theta \, d\theta}{(\dfrac{3}{2} \sec \theta)(3 \tan \theta)} = \int \dfrac{1}{3} d\theta = \dfrac{1}{3} \theta + c$.

Since $x = \dfrac{3}{2} \sec \theta$, we have $\sec \theta = \dfrac{2x}{3}$, which implies $\cos \theta = \dfrac{3}{2x}$.

Using $\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\dfrac{4x^2}{9} - 1} = \dfrac{\sqrt{4x^2 - 9}}{3}$, we get $\theta = \sec^{-1}(\dfrac{2x}{3})$.

Alternatively, using $\tan \theta = \dfrac{\sqrt{4x^2 - 9}}{3}$, we have $\theta = \tan^{-1}(\dfrac{\sqrt{4x^2 - 9}}{3})$.

Thus, $I = \dfrac{1}{3} \tan^{-1}(\dfrac{\sqrt{4x^2 - 9}}{3}) + c$.

Answer: $\dfrac{1}{3} \tan ^{-1}\left(\dfrac{\sqrt{4 x^2-9}}{3}\right)+c$

Hence, the correct answer is option: 'C'.

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