Use Taylor series representation for $\arctan$:
$$
\frac{\arctan(x)}{x} = \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} x^{2k}
$$
and integrate term-wise, for $0<x<1$:
$$ \begin{eqnarray}
\int_0^x \frac{\arctan(t)}{t} \mathrm{d} t &=& \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)^2} x^{2k+1} = x \sum_{k=0}^\infty \frac{(-x^2)^{k}}{k!} \frac{\left(\frac{1}{2}\right)_k \cdot \left(\frac{1}{2}\right)_k \cdot k!}{\left(\frac{3}{2}\right)_k \cdot \left(\frac{3}{2}\right)_k} \\ &=& x \cdot {}_3 F_2\left(\frac{1}{2},\frac{1}{2}, 1; \frac{3}{2}, \frac{3}{2}; -x^2 \right)
\end{eqnarray}
$$
wher $(a)_k := a(a+1)\cdots(a+k-1)$ denotes Pochhammer symbol and ${}_pF_q$ denotes generalized hypergeometric function. For $x>1$, we split the integration range $(0,x)$ into $(0,1)$ and $(1,x)$ and perform a change of variables $t \to 1/t$ in the second integral:
$$\begin{eqnarray}
\int_0^x \frac{\arctan(t)}{t} \mathrm{d} t &=& \int_0^1 \frac{\arctan(t)}{t} \mathrm{d} t + \int_1^{x} \frac{\arctan(t)}{t} \mathrm{d} t \\
&=& \int_0^1 \frac{\arctan(t)}{t} \mathrm{d} t + \int_{1/x}^1 \frac{\arctan(1/t)}{t} \mathrm{d} t
\end{eqnarray}
$$
Using the identity $\arctan(t) + \arctan(1/t) = \frac{\pi}{2}$ which holds for $t>0$, we have:
$$\begin{eqnarray}
\int_0^x \frac{\arctan(t)}{t} \mathrm{d} t &=& \int_0^1 \frac{\arctan(t)}{t} \mathrm{d} t + \int_1^{x} \frac{\arctan(t)}{t} \mathrm{d} t \\
&=& \int_0^{1/x} \frac{\arctan(t)}{t} \mathrm{d} t + \frac{\pi}{2} \int_{1/x}^1 \frac{1}{t} \mathrm{d} t \\
&=& \frac{\pi}{2} \ln(x) + \frac{1}{x} \cdot {}_3 F_2\left(\frac{1}{2},\frac{1}{2}, 1; \frac{3}{2}, \frac{3}{2}; -\frac{1}{x^2} \right)
\end{eqnarray}
$$
As a corollary, $\int_0^x \frac{\arctan(t)}{t} \mathrm{d}t$ diverges logarithmically as $x$ grows large. Of course, this can be determined by much simpler means, as in Gerry's answer.
