求定积分:∫(0-2) dx/[根号下(x+1)+根号下(x+1)^3,帮我解决一下,谢谢!

令t=(x+1)^(1/2)
所以x=t^2-1
所以dx=2tdt
当x=0,t=1,x=2,t=3^(1/2)
原式=∫2tdt/(t+t^3)].上限3^(1/2).下限1
=2∫dt/(1+t^2)
=2[arctant].上限3^(1/2).下限1
=2[arctan3^(1/2)-arctan1]
=2(π/3-π/4)
=π/6