The idea is to relate this expression to the known power series expansion
#1/(1-x)=sum_(n=0)^oox^n#
Temporarily disregard the #x^2# and consider
#f(x)=x^2 1/(1-2x)^2#.
Take the integral of #1/(1-2x)^2#:
#intdx/(1-2x)^2#
Quick substitution:
#u=1-2x#
#du=-2dx, -1/2du=dx#
#-1/2intu^-2du=1/(2u)=1/(2(1-2x))#
Thus, knowing that differentiating this integrated expression returns the original #1/(1-2x)^2,# we can say
#f(x)=x^2d/dx1/(2(1-2x))#
Let's find the power series representation for the differentiated expression:
#f(x)=x^2d/dx1/2*1/(1-2x)#
We can easily relate #1/(1-2x)# to #1/(1-x)=sum_(n=0)^oox^n#:
#1/(1-2x)=sum_(n=0)^oo(2x)^n=sum_(n=0)^oo2^nx^n#
So,
#f(x)=x^2d/dx1/2sum_(n=0)^oo2^nx^n#
We can absorb the #1/2# in:
#f(x)=x^2d/dxsum_(n=0)^oo2^(n-1)x^n#
Differentiate the summation with respect to #x#, recalling that differentiating the summation causes the index to shift up by #1#:
#f(x)=x^2sum_(n=0)^ood/dx2^(n-1)x^n#
#f(x)=x^2sum_(n=1)^oo2^(n-1)nx^(n-1)#
Multiply in the #x^2:#
#f(x)=sum_(n=1)^oo2^(n-1)nx^(n-1+2)#
#f(x)=sum_(n=1)^oo2^(n-1)nx^(n+1)#