Let u=1+2x, then x=(u-1)/2 and dx=du/2. The integrand becomes (u-1)/(2u³))=½u⁻²-½u⁻³ and dx=½du.
So ¼∫(u⁻²-u⁻³)du=¼[-1/u+1/(2u²)]=¼[-1/(1+2x)+1/(2(1+2x)²)]. Not forgetting the constant of integration:
¼[-1/(1+2x)+1/(2(1+2x)²)]+C=-(1+4x)/(8(1+2x)²)+C.