Question

帮忙求解吧。

Answer 1

令x=2tant
原式=∫2sect/2tant*2sec²tdt
=2∫sec^3 tdt
=2*∫sectd(tant)
=2sect*tant-2*∫sect*tan²tdt
=2sect*tant-2∫[sec^3 t-sect]dt
=2sect*tant-2*∫sec^3 tdt+2*∫sectdt
所以原式=sect*tant+∫sectdt
=sect*tant+ln|sect+tant|+C
=x/2*√(x²+4)/2+ln[x/2+√(x²+4)/2]+C1
=x*√(x²+4)/4+ln|x+√(x²+4)|+C

追问

麻烦解答

追答

搞错了，不好意思
令x=2tant
原式=∫2sect/2tant*2sec²tdt
=2∫1/(sint*cos²t)dt
=2*∫(sin²t+cos²t)/(sint*cos²t)dt
=2∫sint/cos²t+2*∫csctdt
=-2∫dcost/cos²t+2ln|csct-cott|
=2/cost+2ln|csct-cott|+C
=√(x²+4)+2ln|(√(x²+4)-2)/x|+C

Answer 2

三角换元