Question

用第二类换元法求积分 1/x^2*根号（4+x^2)

Answer 1

∫1/x^2*√(4+x^2)dx
=∫1/(x/2)^2*√(1+(x/2)^2)dx/2
x/2=tant
=∫1/tan^2t*√(1+tan^2t)dtant
=∫sec^3t/tan^2tdt
=∫sect*csc^2tdt
=-∫sectdcott
=-sectcott+∫cottdsect
=-√(1+tan^2t)/tant+∫cotttantsectdt
=-√(4+x^2)/x+∫sectdt
=-√(4+x^2)/x+∫secx(tanx+sect)/(sect+tant)dt
=-√(4+x^2)/x+∫(secxtanx+sec^2t)/(sect+tant)dt
=-√(4+x^2)/x+∫1/(sect+tant)d(sect+tant)
=-√(4+x^2)/x+ln(sect+tant)
=-√(4+x^2)/x+∫1/costdt
=-√(4+x^2)/x+∫cost/cos^2tdt
=-√(4+x^2)/x+ln(x+√(4+x^2))+C