Answer
See below
Work Step by Step
According to Gram-Schmidt process, we obtain:
$v_1=p_1(x)=1+x^2\\
v_2=p_2(x)-\frac{(p_2(x),v_1)}{||v_1||^2}v_1\\
=2-x+x^3-\frac{((2-x+x^3),(1+x^2))}{||1+x^2||^2}(1+x^2)\\
=2-x+x^3-\frac{2.1+(-1)0+0.1+1.0}{1^2+0^2+1^2} (1+x^2) \\
=2-x+x^3-(1+x^2)\\
=1-x-x^2+x^3\\
v_3=p_3(x)-\frac{(p_3(x),v_1)}{||v_1||^2}v_1-\frac{(p_3(x),v_2)}{||v_2||^2}v_2\\
=-x+2x^2-\frac{((-x+2x^2),(1+x^2))}{||1+x^2||^2}(1+x^2)-\frac{((-x+2x^2),(1-x-x^2+x^3))}{||1-x-x^2+x^3||^2}(1-x-x^2+x^3)\\
=-x+2x^2-\frac{.1+(-1).0+2.1}{1^2+0^2+1^2} (1+x^2) -\frac{0.1+(-1).(-1)+2.(-1)+0.1}{1^2+(-1)^2+(-1)^2+1^2}(1-x-x^2+x^3)\\
=-x+2x^2-(1+x^2)+\frac{1}{4}(1-x-x^2+x^3)\\
=\frac{1}{4}(-3-5x+3x^2+x^3)$
Hence, a corresponding orthonormal set of vector is:
$\{1+x^2,1-x-x^2+x^3,\frac{1}{4}(-3-5x+3x^2+x^3)\}$
