Question

帮忙求不定积分

(sinx+cosx)/(sinx-cosx)
[(1+x)/(1-x)]^1/2

Answer 1

Answer 2

∫(sinx+cosx)/(sinx-cosx) dx
=∫(sinx+cosx)^2/[(sinx)^2-(cosx)^2] dx
=-∫ (1+ sin2x)/cos2x dx
=-∫ ( sec2x + tan2x) dx
= - [(1/2)ln|sec2x+ tan2x| - (1/2)ln|cos2x| ) + C
=(1/2)ln |cos2x/(sec2x+tan2x)| + C

∫√ [(1+x)/(1-x)] dx
=∫ [(1+x)/√(1-x^2)] dx
let
x= siny
dx = cosy dy
∫ [(1+x)/√(1-x^2)] dx
=∫ (1+siny) dy
= y - cosy + C
= arcsinx -√(1-x^2) + C

Answer 3