Question

定积分这个怎么计算？

需要详细过程

Answer 1

{ln[x+√(1+x^2)]}'
= [x+√(1+x^2)]' / [x+√(1+x^2)]
= [1+x/√(1+x^2)] / [x+√(1+x^2)]
= [√(1+x^2)+x] / {[x+√(1+x^2)]√(1+x^2)}
= 1 / √(1+x^2) 代入即得