Consider the geometric series:
∞∑n=0qn=11−q
for |q|<1.
Let q=−x2
11+x2=∞∑n=0(−1)nx2n
Integrate term by term:
∫x0dt1+t2=∞∑n=0(−1)n∫x0t2ndt
arctanx=∞∑n=0(−1)nx2n+12n+1dt
Integrate again, considering that integrating by parts:
∫arctanxdx=xarctanx−∫x1+x2dx
∫arctanxdx=xarctanx−12∫d(1+x2)1+x2
∫arctanxdx=xarctanx−12ln(1+x2)+c
so:
∫x0arctant=∞∑n=0(−1)n2n+1∫x0t2n+1dt
xarctanx−12ln(1+x2)=∞∑n=0(−1)nx2n+2(2n+1)(2n+2)
Let now x=1√3. As 0<1√3<1 the series is convergent in this point:
1√3arctan(1√3)−12ln(1+13)=∞∑n=0(−1)n1(√3)2n+21(2n+1)(2n+2)
note that:
(√3)2n+2=3⋅(√3)2n=3⋅3n
so:
1√3arctan(1√3)−12ln(1+13)=13∞∑n=0(−1)n13n1(2n+1)(2n+2)
Extract now the first two terms for n=0 and n=1:
1√3arctan(1√3)−12ln(1+13)=16−1108+13∞∑n=2(−1)n13n1(2n+1)(2n+2)
1√3arctan(1√3)−12ln(1+13)−16+1108=13∞∑n=2(−1)n13n1(2n+1)(2n+2)
√3arctan(1√3)−32ln(43)−1736=∞∑n=2(−1)n13n1(2n+1)(2n+2)