∫ 1/[(ax + b)x] dx
= (1/b)∫ [(ax + b) - ax]/[(ax + b)x] dx
= (1/b)∫ 1/x dx - (a/b)∫ dx/(ax + b)
= (1/b)∫ 1/x dx - (1/b)∫ d(ax + b)/(ax + b)
= (1/b)ln|x| - (1/b)ln|ax + b| + C
= (1/b)ln|x/(ax + b)| + C