Answer
$\{1+2x,-1+x+x^2\}$
Work Step by Step
We are given:
$f_1(x)=1+2x\\
f_2(x)=-2-x+x^2$
According to Gram-Schmidt we have:
$v_1=f_1(x)=1+2x\\
v_2=f_2(x)-\frac{(f_2(x),v_1)}{||v_1||^2}v_1\\
=-2-x+x^2-\frac{(-2-x+x^2,1+2x)}{||1+2x||^2}(1+2x) \\
=-2-x+x^2-\frac{\int ^1_0 (-2-x+x^2)(1+2x)dx}{\int^1_0(1+2x)^2dx}(1+2x) \\
=-2-x+x^2-\frac{\int ^1_0 (-2-4x-x-2x^2+x^2+2x^3)dx}{\int^1_0(1+4x+4x^2)dx}(1+2x) \\
=-2-x+x^2-\frac{\int ^1_0 (-2-5x-x^2+2x^3)dx}{x+2x^2+\frac{4}{3}x^3|^1_0}(1+2x) \\
=-2-x+x^2-\frac{(-2x-\frac{5}{2}x^2-\frac{1}{3}x^3+\frac{1}{2}x^4)|^1_0}{x+2+\frac{4}{3}}(1+2x) \\
=-2-x+x^2-\frac{(-2-\frac{5}{2}-\frac{1}{3}+\frac{1}{2})}{\frac{13}{3}}(1+2x) \\
=-2-x+x^2+1+2x\\
=-1+x+x^2$
Hence, an orthogonal basic spanned by the given factors is:
$\{1+2x,-1+x+x^2\}$
