Question

第二积分换元，经过变量代换x=tant,∫√(1+x²)dx怎么算？

Answer 1

x=tant
dx=(sect)^2 dt
∫√(1+x^2)dx
=∫ (sect)^3 dt
=∫ sect. dtant
=sect.tant - ∫ sect.(tant)^2 dt
=sect.tant - ∫ sect.[(sect)^2-1] dt
2∫ (sect)^3 dt = sect.tant + ∫ sect dt
∫ (sect)^3 dt
= (1/2)[ sect.tant + ln|sect+ tant| ] +C
//
∫√(1+x^2)dx
=∫ (sect)^3 dt
= (1/2)[ sect.tant + ln|sect+ tant| ] +C
= (1/2)[ x.√(1+x^2) + ln|√(1+x^2)+ x| ] +C