Question

x根号x+2不定积分,就是x√x+2积分,

Answer 1

答：

∫x√(x+2)dx

=∫(x+2-2)√(x+2)dx

=∫(x+2)^(3/2)d(x+2)-2∫(x+2)^(1/2)d(x+2)

=(2/5)(x+2)^(5/2)-2*(2/3)(x+2)^(3/2)+C

=(2/5)(x+2)^(5/2)-(4/3)(x+2)^(3/2)+C