Q88 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics

$\int \dfrac{x^2\, \text{d}x}{(x^2 + 2)(x^2 + 5)} =$

  • A
    $-\left(\dfrac{\sqrt{2}}{3}\tan^{-1}\dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{5}}{3}\tan^{-1}\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration
  • B
    $\left(\dfrac{\sqrt{2}}{3}\tan^{-1}\dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{5}}{3}\tan^{-1}\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration
  • C
    $\dfrac{\sqrt{2}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{2}}\right) - \dfrac{\sqrt{5}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration
  • Correct answer
    D
    $-\dfrac{\sqrt{2}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + \dfrac{\sqrt{5}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration

Solution

Let $I = \int \dfrac{x^2}{(x^2 + 2)(x^2 + 5)} \, \text{d}x$

To resolve into partial fractions, substitute $x^2 = t$:

$\dfrac{t}{(t + 2)(t + 5)} = \dfrac{A}{t + 2} + \dfrac{B}{t + 5}$

$t = A(t + 5) + B(t + 2)$

Putting $t = -2$, we get $-2 = 3A \Rightarrow A = -\dfrac{2}{3}$

Putting $t = -5$, we get $-5 = -3B \Rightarrow B = \dfrac{5}{3}$

Therefore, $\dfrac{x^2}{(x^2 + 2)(x^2 + 5)} = -\dfrac{2}{3(x^2 + 2)} + \dfrac{5}{3(x^2 + 5)}$

Integrating with respect to $x$:

$I = -\dfrac{2}{3} \int \dfrac{1}{x^2 + (\sqrt{2})^2} \, \text{d}x + \dfrac{5}{3} \int \dfrac{1}{x^2 + (\sqrt{5})^2} \, \text{d}x$

$I = -\dfrac{2}{3} \left( \dfrac{1}{\sqrt{2}} \tan^{-1} \dfrac{x}{\sqrt{2}} \right) + \dfrac{5}{3} \left( \dfrac{1}{\sqrt{5}} \tan^{-1} \dfrac{x}{\sqrt{5}} \right) + c$

$I = -\dfrac{\sqrt{2}}{3} \tan^{-1} \left( \dfrac{x}{\sqrt{2}} \right) + \dfrac{\sqrt{5}}{3} \tan^{-1} \left( \dfrac{x}{\sqrt{5}} \right) + c$

Answer: $-\dfrac{\sqrt{2}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + \dfrac{\sqrt{5}}{3}\tan^{-1}\left(\dfrac{x}{\sqrt{5}}\right) + c$, where c is the constant of integration

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

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