Q73 / 150 · MHT CET 2025 (23 Apr Shift 1) · Mathematics

$\int \frac{\mathrm{d} x}{\sqrt{x}+x}=$

  • A
    $\log \sqrt{x}+c, \quad$ where $c$ is the constant of integration.
  • B
    $\log (\sqrt{x}+x)+c, \quad$ where $c$ is the constant of integration.
  • C
    $\log (1+\sqrt{x})+c, \quad$ where $c$ is the constant of integration.
  • Correct answer
    D
    $2 \log (1+\sqrt{x})+c, \quad$ where $c$ is the constant of integration.

Solution

Let $u = \sqrt{x}$, so $x = u^2$ and $\mathrm{d}x = 2u\,\mathrm{d}u$. Substituting into the integral gives

$I = \int \frac{2u\,\mathrm{d}u}{u + u^2} = \int \frac{2u\,\mathrm{d}u}{u(1 + u)}$

For $u > 0$ (since $x > 0$), this simplifies to

$I = 2\int \frac{\mathrm{d}u}{1 + u} = 2\ln|1 + u| + C$

Since $\sqrt{x} \geq 0$, we have $|1 + \sqrt{x}| = 1 + \sqrt{x}$ and thus

$I = 2\ln(1 + \sqrt{x}) + C$

This result corresponds to option D.

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

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