y'=[√(x^2+1)-x/√(x^2+1)]/(x^2+1)=[x^2+1-x]/[(x^2+1)√(x^2+1)]
微分dy=[x^2+1-x]/[(x^2+1)√(x^2+1)]dx答案是(x^2+1)^-3/2 dx答案是(x^2+1)^-3/2 dx哦，我中间有一步算错了，少算了(x^2+1)'=2x了。y'=[√(x^2+1)-x/2√(x^2+1)*2x]/(x^2+1)=[x^2+1-x^2]/[(x^2+1)√(x^2+1)]=1/[(x^2+1)√(x^2+1)]微分dy=1/[(x^2+1)√(x^2+1)]dx哦，谢谢你啦