Approximate an integral with a square root

Enter lower limit $0$, upper limit $4$ and integrand $2/(2+\sqrt x)$ in the calculator integral template. The value is approximately $2.454822556$, hence $2.45$ to three significant figures, as printed.

Independently, put $t=\sqrt x$, so $dx=2t\,dt$ and the limits are $0,2$. Then $\int_0^2\frac{4t}{2+t}\,dt=\int_0^2(4-\frac8{t+2})\,dt=[4t-8\ln(t+2)]_0^2=8-8\ln2$, agreeing with the printed decimal.