integration by parts we get

\(\displaystyle -\int_0^1 \frac{\tan ^{-1}x}{x} \, dx\)

\(\displaystyle -\int_0^1 \left(\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k}}{(2k+1)}\right) \, dx\)

\(\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)}\int_0^1 x^{2k} \, dx\)

\(\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)^2}\)

\(\displaystyle -Catalan\) constant

$-0.915966$