求定积分：[0,2]∫ dx / [√(x+1)+√(x+1)³]
令√(x+1)=u,则x=u²-1,dx=2udu,x=0时u=1；x=2时u=√3；代入原式得：
[0,2]∫ dx / [√(x+1)+√(x+1)³]=[1,√3]∫2udu/(u+u³)=[1,√3]2∫du/(1+u²)=2arctanu︱[1,√3]
=2[arctan√3-arctan1]=2(π/3-π/4)=π/6.
求定积分：∫ dx / [√(x+1)+√(x+1)^3]
求定积分：[0,2]∫ dx / [√(x+1)+√(x+1)³]
令√(x+1)=u,则x=u²-1,dx=2udu,x=0时u=1；x=2时u=√3；代入原式得：
[0,2]∫ dx / [√(x+1)+√(x+1)³]=[1,√3]∫2udu/(u+u³)=[1,√3]2∫du/(1+u²)=2arctanu︱[1,√3]
=2[arctan√3-arctan1]=2(π/3-π/4)=π/6.