Evaluate $\int \frac{x}{\sqrt{x} + 1} dx$.

$2 \left[ \frac{x\sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - \log |(\sqrt{x} + 1)| \right] + C$

The correct answer is Option (3) → $2 \left[ \frac{x\sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - \log |(\sqrt{x} + 1)| \right] + C$

Let $I = \int \frac{x}{\sqrt{x} + 1} dx$

Put  $\sqrt{x} = t \Rightarrow \frac{1}{2\sqrt{x}} dx = dt$

$\Rightarrow dx = 2\sqrt{x} dt$

$∴I = 2 \int \left( \frac{x \sqrt{x}}{t + 1} \right) dt = 2 \int \frac{t^2 \cdot t}{t + 1} dt = 2 \int \frac{t^3}{t + 1} dt$

$= 2 \int \frac{t^3 + 1 - 1}{t + 1} dt = 2 \int \frac{(t+1)(t^2 - t + 1)}{t + 1} dt - 2 \int \frac{1}{t + 1} dt$

$[∵a^3 + b^3 = (a+b)(a^2 - ab + b^2)]$

$= 2 \int (t^2 - t + 1) dt - 2 \int \frac{1}{t + 1} dt$

$= 2 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log |(t + 1)| \right] + C$

Put $t = \sqrt{x}$

$= 2 \left[ \frac{x\sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - \log |(\sqrt{x} + 1)| \right] + C$