∫ x√(ax + b) dx
= (1/a)∫ [(ax + b) - b]√(ax + b) dx
= (1/a)∫ (ax + b)^(3/2) dx - (b/a)∫ √(ax + b) dx
= (1/a)(1/a)∫ (ax + b)^(3/2) d(ax + b) - (b/a)(1/a)∫ √(ax + b) d(ax + b)
= (1/a²)(2/5)(ax + b)^(5/2) - (b/a²)(2/3)(ax + b)^(3/2) + C
= [2/(15a²)](3ax - 2b)(ax + b)^(3/2) + C