用变量代换法求§[x(x+2)^1/2]dx不定积分

令(x+2)^1/2=t
x=t^2-2
dx=2tdt
§[x(x+2)^1/2]dx
=∫(t^2-2)t*2tdt
=∫(2t^4-4t^2)dt
=2t^5/5-4t^3/3+c
=2/5*(x+2)^(5/2)-4/3*(x+2)^(3/2)+c