Question

求√1-t的不定积分dt

Answer 1

问：求√1-t的不定积分dt

答：∫dt /(1+√(1+t) )lett = (tany)^2dt = 2tany (secy)^2 dy∫dt /(1+√(1+t) )=∫2tany (secy)^2 /(1+secy) dy=2∫ siny/[(cosy)^2(1+cosy)] dyletz= cosy

答：dz = siny dy2∫ siny/[(cosy)^2(1+cosy)] dy=2∫ dz/[z^2(1+z)]let1/[z^2(1+z)] = a1/z+ a2/z^2+ b/(1+z)=>1= a1z(1+z) + a2(1+z) +bz^2z=0, a2=1z=-1, b=1coef. of z^2a1+b=0a1=-1

答：1/[z^2(1+z)] =-1/z+ 1/z^2+ 1/(1+z)2∫ dz/[z^2(1+z)]=2∫ [-1/z+ 1/z^2+ 1/(1+z)]dz=2(ln|(1+z)/z| - 1/z ) + Cwhere z = 1/√(1+t)t = (tany)^2tany = √tz= cosy =1/√(1+t)