integration

y = 2x + 1
dy = 2dx
∫x√(2x+1)dx
= (1/4)∫(y - 1)√y dy
= (1/4)∫[ y^(3/2) - y^(1/2) ] dy
= (1/4) [ (2/5)y^(5/2) - (2/3)y^(3/2) ] + C, C is constant
= (1/2) [ y^(5/2) / 5 - y^(3/2) / 3 ] + C
= (1/30) [ 3y^(5/2) - 5y^(3/2) ] + C
= [ 3y^(5/2) - 5y^(3/2) ] / 30 + C
= [ 3 (2x + 1)^(5/2) - 5 (2x + 1)^(3/2) ] / 30 + C

2007-10-12 20:55:35 補充：
約簡 d, = [ 3 (2x + 1) - 5 ] (2x + 1)^(3/2) / 30 + C= (6x - 2) (2x + 1)^(3/2) / 30 + C= (3x - 1) (2x + 1)^(3/2) / 15 + C