题目

sqrt(1+x*x)不定积分

提问时间：2021-02-26

答案

令x=tant dx=sec^2tdt
∫√(1+x^2)dx
=∫sec^3tdt
=∫(sin^2t+cos^2t)/cos^3tdt
=∫dt/cost+∫sin^2t/cos^3tdt
=∫sectdt-∫sint/cos^3td(cost)
=∫sectdt-1/2*∫sintd(1/cos^2t)
=∫sectdt-1/2*sint/cos^2t-1/2*∫dt/cost
=1/2*ln|sect+tant|-1/2*sint/cos^2t+C
=x/2*√(1+x^2)+1/2*ln|x+√(1+x^2)|+C