The value of

Let, $\quad \mathrm{I}=\int \frac{x}{\sqrt{x}+1} d x$

Put $\sqrt{x}=t \Rightarrow x=t^{2} \quad \therefore d x=2 t . d t$

So,

$I=\int \frac{t^{2} \cdot 2 t \cdot d t}{t+1}=2 \int \frac{t^{3}}{t+1} d t=2 \int \frac{t^{3}+1-1}{t+1} d t$

$=2 \int \frac{t^{3}+1}{t+1} d t-2 \int \frac{1}{t+1} d t$

$=2 \int \frac{(t+1)\left(t^{2}-t+1\right)}{t+1} d t-2 \int \frac{1}{t+1} d t$

$=2 \int\left(t^{2}-t+1\right) d t-2 \int \frac{1}{t+1} d t$

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

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

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

Therefore,

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