Consider the geometric series:
sum_(n=0)^oo q^n = 1/(1-q)
for absq < 1.
Let q=-x^2
1/(1+x^2) = sum_(n=0)^oo (-1)^nx^(2n)
Integrate term by term:
int_0^x dt/(1+t^2) = sum_(n=0)^oo (-1)^n int_0^x t^(2n)dt
arctanx = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)dt
Integrate again, considering that integrating by parts:
int arctanx dx = xarctanx - int x/(1+x^2)dx
int arctanx dx = xarctanx - 1/2 int (d(1+x^2))/(1+x^2)
int arctanx dx = xarctanx - 1/2 ln(1+x^2)+c
so:
int_0^x arctant = sum_(n=0)^oo (-1)^n/(2n+1) int_0^x t^(2n+1)dt
xarctanx - 1/2 ln(1+x^2) = sum_(n=0)^oo (-1)^n x^(2n+2)/((2n+1)(2n+2))
Let now x=1/sqrt3. As 0 < 1/sqrt3 < 1 the series is convergent in this point:
1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) = sum_(n=0)^oo (-1)^n 1/(sqrt3)^(2n+2)1/((2n+1)(2n+2))
note that:
(sqrt3)^(2n+2) = 3*(sqrt3)^(2n) = 3*3^n
so:
1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) = 1/3sum_(n=0)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))
Extract now the first two terms for n=0 and n=1:
1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) = 1/6 - 1/108 +1/3sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))
1/sqrt3arctan(1/sqrt3) - 1/2 ln(1+1/3) - 1/6 + 1/108 =1/3sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))
sqrt3arctan(1/sqrt3) - 3/2 ln(4/3) - 17/36 =sum_(n=2)^oo (-1)^n 1/3^n 1/((2n+1)(2n+2))