\(\displaystyle\int_0^1x\,\mathrm dx\) corresponds to the area under the curve \(y=x\) from 0 to 1. This region is a right triangle with base and height 1, so the area of the triangle, and thus the value of the integral, is \(\dfrac12\times1\times1=\dfrac12\).
\(\displaystyle\int_0^1\sqrt{1-x^2}\,\mathrm dx\) is the area of one quadrant of the unit circle. Recall that \(x^2+y^2=1\) is the equation of the circle with radius 1 centered at the origin. Solving for \(y\) gives \(y=\pm\sqrt{1-x^2}\), where the positive root corresponds to the top half of the circle. The top half is defined over the interval \([-1,1]\), but you're only interested in the right half, \([0,1]\). So this area is \(\dfrac14\pi\times1^2=\dfrac\pi4\).
Put these together to get
\(\displaystyle\int_0^1x\,\mathrm dx+\int_0^1\sqrt{1-x^2}\,\mathrm dx=\int_0^1(x+\sqrt{1-x^2})\,\mathrm dx=\frac12+\frac\pi4\)
