2.2.51 ** Find the value of the constant $c$ such that the function $f(x)=\arctan \frac{2-2 x}{1+4 x}+c$ is an odd function on the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$.
Assuming the constant $c$ exists, since the function $f(x)$ is an odd function, we have $f(0)=\arctan 2 + c=0$, so the only possible value for $c$ is $-\arctan 2$.

Below, we prove that in the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$, the function $f(x)=\arctan \frac{2-2 x}{1+4 x}-\arctan 2$ is an odd function, i.e., it satisfies the relation: $f(x)=-f(-x)$.

Let $z=\frac{2-2 x}{1+4 x}$, it is easy to see that $z=-\frac{1}{2}+\frac{5}{2(1+4 x)}$. When $-\frac{1}{4}<x<\frac{1}{4}$, we have $z>0$. When $x=0$, $z=2$; when $x=\frac{1}{4}$, $z=\frac{3}{4}$. Therefore, $\arctan \frac{3}{4}<\arctan z<\frac{\pi}{2}, \arctan \frac{3}{4}-\arctan 2<\arctan z-\arctan 2<\frac{\pi}{2}-$ $\arctan 2$. Hence, (1) is equivalent to
$$
\tan f(x)=\tan (-f(-x)).
$$

But by the trigonometric formula, we get $\tan f(x)=-2 x, \tan [-f(-x)]=-2 x$, so (2) holds, and thus (1) is established. Therefore, $c-\cdots \arctan 2$.