Start from:
#arctan(4x) = int_0^(4x) dt/(1+t^2)#
Expand now the integrand using the geometric series:
#arctan(4x) = int_0^(4x) sum_(n=0)^oo (-1)^nt^(2n)dt#
for #abs (4x) < 1# the series is absolutely convergent and we can integrate term by term:
#arctan(4x) = sum_(n=0)^oo (-1)^nint_0^(4x) t^(2n)dt#
#arctan(4x) = sum_(n=0)^oo (-1)^n (4x)^(2n+1)/(2n+1)#
#arctan(4x) = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)x^(2n+1)#
Dividing by #x^(-7)# means multiplying by #x^7# and we can do it again term by term:
#arctan(4x)/x^(-7) = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)x^(2n+8)#
and integrating again term by term:
#int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)int x^(2n+8) dx#
#int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/((2n+1)(2n+9) )x^(2n+9) +C#
