∵√[x(x+1)³]=x²*√[(x+1)³/x³]∴原式=∫[0,1]1/x²*√[x³/(x+1)³]*dx令√[x³/(x+1)³]=t³,则x/(x+1)=t²,x=t²/(1-t²),dx=2tdt/(1-t²)²当x从0变化到1时,t从0变化到√2/2原式=∫[0,√2/2]2dt=2(√2/2-0)=√2
令x=tanu
