Note that:
#int dx/(2-3x)^2 = -1/3 int (2-3x)^-2 d(2-3x)#
#int dx/(2-3x)^2 = -1/3 (2-3x)^-1/(-1) +C#
#int dx/(2-3x)^2 = 1/3 1/(2-3x) +C#
So that:
#1/3 d/dx (1/(2-3x)) = 1/(2-3x)^2#
Now express:
#1/(2-3x) = 1/2 1/(1-((3x)/2)#
We know that a geometric series or ratio #abs q < 1# is convergent and:
#sum_(n=0)^oo q^n = 1/(1-q)#
so with #q=(3x)/2#
we have:
# sum_(n=0)^oo ((3x)/2)^n = 1/(1-((3x)/2))#
converging absolutely for #abs((3x)/2) < 1#, that is for #x in (-2/3,2/3)#, then:
#1/(2-3x)^2 = 1/3 d/dx (1/(2-3x)) = 1/6 d/dx(sum_(n=0)^oo ((3x)/2)^n)#
In the interval of convergence we can differentiate the series term by term:
#1/(2-3x)^2 = 1/6 sum_(n=0)^oo d/dx((3x)/2)^n#
#1/(2-3x)^2 = 1/6 sum_(n=0)^oo (3n)/2((3x)/2)^(n-1)#
The term for #n=0# is null, so:
#1/(2-3x)^2 = 1/4 sum_(n=1)^oo n((3x)/2)^(n-1)#
and changing index: #k=n-1#:
#1/(2-3x)^2 = 1/4 sum_(k=0)^oo (3^k(k+1))/2^k x^k#
Or alternatively:
#1/(2-3x)^2 = sum_(k=0)^oo (3^k(k+1))/2^(k+2) x^k#
converging for #abs x < 2/3#
