Not $\zeta(3)$

This is the Dirichlet Beta Function evaluated at 3.

For an elementary proof, see this ....

Consider the polylogarithm function $\text{Li}_s (z)$ :

$\begin{aligned} \text{Li}_s (z) & = \sum_{k=1}^\infty \frac {z^k}{k^s} = \frac z{1^s} + \frac {z^2}{2^s} + \frac {z^3}{3^s} + \frac {z^4}{4^s} + \cdots \\ \text{Li}_3 (i) & = \frac i{1^3} + \frac {i^2}{2^3} + \frac {i^3}{3^3} + \frac {i^4}{4^3} + \frac {i^5}{5^3} + \frac {i^6}{6^3} + \frac {i^7}{7^3} + \cdots & \small \color{#3D99F6} \text{where }i = \sqrt{-1} \text{ denotes the imaginary unit.} \\ \text{Li}_3 (i) & = i - \frac 1{2^3} - \frac i{3^3} + \frac 1{4^3} + \frac i{5^3} - \frac 1{6^3} - \frac i{7^3} + \cdots \\ \implies S & = \Im \left(\text{Li}_3(i) \right) & \small \color{#3D99F6} \text{where }\Im (\cdot) \text{ denotes the imaginary part function.} \\ & \approx \Im \left(-0.112693 + 0.968946i \right) \\ & = \boxed{0.968946} \end{aligned}$

The above expression or $S$ can be written as $\int_{0}^1\frac{1}{x}(\int \frac{tan^{-1}(x)}{x} dx) dx$ =0.968946