Example 5 Find the constant $c$, such that $f(x)=\arctan \frac{2-2 x}{1+4 x}+c$ is an odd function in the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$.
If $f(x)$ is an odd function in $\left(-\frac{1}{4}, \frac{1}{4}\right)$, then the necessary condition is $f(0)=0$, i.e., $c=$ $-\arctan 2$. When $c=-\arctan 2$, $\tan \left(\arctan \frac{2-2 x}{1+4 x}-\arctan 2\right)=\frac{\frac{2-2 x}{1+4 x}-2}{1+\frac{2-2 x}{1+4 x} \cdot 2}$ $=\frac{2-2 x-2-8 x}{1+4 x+4-4 x}=-2 x$, i.e., $f(x)=\arctan (-2 x) . f(-x)=\arctan (-(-2 x))$ $=\arctan 2 x=-f(x)$, hence $f(x)$ is an odd function in $\left(-\frac{1}{4}, \frac{1}{4}\right)$.