Start from the sum of the geometric series:
#sum_(k=0)^oo q^k = 1/(1-q)#
For #abs(q) <1#
Let #q=3x^2# so that we have:
#sum_(k=0)^oo 3^kx^(2k) = 1/(1-3x^2)#
and multiplying by #6x#:
#6sum_(k=0)^oo 3^kx^(2k+1) = (6x)/(1-3x^2)#
The series is convergent for #3x^2 < 1#, that is for #x in (-1/sqrt3,1/sqrt3)#. Inside this interval we can integrate the series term by term and obtain a series with the same radius of convergence.
#6sum_(k=0)^oo 3^kint_0^x t^(2k+1)dt = int_0^x (6tdt)/(1-3t^2)#
#6sum_(k=0)^oo 3^kx^(2k+2)/(2k+2) = int_0^x -(d(1-3t^2))/(1-3t^2)#
#6sum_(k=0)^oo 3^kx^(2k+2)/(2k+2) = -lnabs(1-3x^2)#
and finally:
#ln(1-3x^2) = -sum_(k=0)^oo 3^(k+1)x^(2k+2)/(k+1)#
