Answer
$$\operatorname{proj}_{{g}} {f}
= -\frac{4}{35}(2x-1).$$
Work Step by Step
Let $f(x)=x^3-x, \quad g(x)=2x-1, \quad C[-1,1] , \quad\langle f, g\rangle=\int_{-1}^1 f(x)g(x) d x.$
We have
\begin{aligned}\langle f, g\rangle&=\int_{-1}^1 (2x^4-x^3-2x^2+x) d x\\
&=\left[\frac{ 2}{5 } x^5-\frac{1}{4}x^4-\frac{2}{3}x^3+\frac{1}{2}x^2\right]_{-1}^1=-\frac{8}{15},\end{aligned}
$$\langle g,g\rangle=\int_{-1}^{1} (4x^2-4x+1) d x= \left[\frac{4}{3}x^3-2x^2+x\right]_{-1}^{1}=\frac{14}{3}.$$
Now, the orthogonal projection of $f$ onto $g$ is given by
$$\operatorname{proj}_{{g}} {f}
=\frac{\langle{f}, {g}\rangle}{\langle{g}, {g}\rangle} {g}=-\frac{4}{35}(2x-1).$$
