\(\int \frac{\sqrt{x}}{x+1} \mathrm{~d} x=\)

\(\begin{aligned} \mathrm{I} & =\int \frac{\sqrt{x}}{x+1} \mathrm{~d} x \\ \text { Let } & \sqrt{x}=\mathrm{t} \\ \Rightarrow & x=\mathrm{t}^2 \\ \mathrm{~d} x & =2 \mathrm{t} \mathrm{dt} \\ \therefore \quad \mathrm{I} & =\int \frac{\mathrm{t}}{\mathrm{t}^2+1} 2 \mathrm{tdt} \\ & =\int \frac{2 \mathrm{t}^2}{\mathrm{t}^2+1} \mathrm{dt} \\ & =2 \int \frac{\mathrm{t}^2}{\mathrm{t}^2+1} \mathrm{dt} \\ & =2 \int\left(\frac{\mathrm{t}^2+1-1}{\mathrm{t}^2+1}\right) \mathrm{dt} \\ & =2\left[\int \frac{\mathrm{t}^2+1}{\mathrm{t}^2+1} \mathrm{dt}-\int \frac{1}{\mathrm{t}^2+1} \mathrm{dt}\right] \\ & =2\left[\mathrm{t}-\tan ^{-1} \mathrm{t}\right]+\mathrm{c} \\ \mathrm{I} & =2\left[\sqrt{x}-\tan ^{-1} \sqrt{x}\right]+\mathrm{c}\end{aligned}\)