Question

∫1/√(1+x²）dx

答案是ln│x+√1+x²│。。。我要过程

Answer 1

设
x=tant
t=arctant
dx=sec^
2t
原式=∫1/sect*sec^2tdt
=∫sectdt
=∫1/costdt
〖∫cscxdx=∫1/sinxdx=∫dx/(2sin(x/2)cos(x/2))=∫d(x/2)/(tan(x/2)cos(x/2))=∫d(tan(x/2)/tan(x/2))=Ln|tan(x/2)|+C
tan(x/2)=sin(x/2)/cos(1/2)=(1-cosx)/sinx=cssx-cotx〗
=∫css(t+派/2)d(t+派/2)
=Ln(csc(t+派/2)-cot(t+派/2))
=Ln|sect+tant|+C
=Ln|sec(arctant)+tanarctant|+C
〖sec(arctant)利用三角关系
tant=x/1直角三角形两条直角边是x和1，斜边是√(x^2+1)则sec(arctant)=√(x^2+1)〗
=Ln(x+√(x^2+1))+C