Answer
$\text{See explanation.}$
Work Step by Step
$\text{The given function is}$
\begin{align}
F(x) = \int_0^x \frac{1}{1+t^2} \ dt + \int_0^{\frac{1}{x}} \frac{1}{1+t^2} \ dt
\end{align}
$\text{Let us differentiate the given function:}$
\begin{align}
F'(x) = \left(\arctan x + \arctan {\frac{1}{x}} \right)'
\end{align}
$\text{We know that the derivative of $\arctan x$ is $\frac{1}{1+x^2}$}$.
$\text{So let us find the derivative of $\arctan {\frac{1}{x}}$ by considering u=$\frac{1}{x} $:}$
\begin{align}
u = \frac{1}{x} \Rrightarrow du = -\frac{1}{x^2} \ dx \ and \ u^2 = \frac{1}{x^2}
\end{align}
$\text{Therefore:}$
\begin{align}
\left(\arctan {\frac{1}{x}} \right)' = \frac{-\frac{1}{x^2}}{1+\frac{1}{x^2}} = - \frac{1}{1+x^2}
\end{align}
$\text{Finally, we can show that the function F(x) is constant:}$
\begin{align}
F'(x) = \left(\arctan x + \arctan {\frac{1}{x}} \right)' = \frac{1}{1+x^2} - \frac{1}{1+x^2} = 0
\end{align}
$\text{Thus, the F(x) is constant on (0, +$\infty$) because its derivative is zero}$
