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

}

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

Let's put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}dx = dt$

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

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

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

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

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

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