求不定积分=∫√(1+1/x²)dx的原函数

求不定积分=∫√(1+1/x²)dx的原函数

令1/x=tant 则 x=cott
∫√(1+1/x²)dx
=∫√(1+tan²t)dcott
=∫sectdcott
=sectcott-∫cottdsect
=csct-∫cott*tant*sectdt
=csctcott-∫sectdt
=csct-ln|sect+tant|
=√((1/tant)^2+1)-ln|√((tant)^2+1)+tant|
=√(x^2+1)-ln|√(x^(-2)+1)+x^(-1)|