Question

1/x²-1的不定积分怎么求

Answer 1

∫ dx/[x√(1+x2)]， x=tanz，dx=sec2zdz，z∈(π/2，π/2) sinz=x/√(1+x2)，cosz=1/√(1+x2) 原式= ∫ sec2z/tanz*secz] dz = ∫ (1/cosz * cosz/sinz) dz = ∫ cscz dz= ln|cscz - cotz| + C = ln|√(1+x2)/x - 1/x| + C = ln|√(1+x2) - 1| - ln|x| + C

追问

∫1/(x²+1)dx

Answer 2

∫dx/x²-1=1/2∫（1/x-1+1/x+1）dx=1/2（ln丨x-1丨-ln丨x+1丨）+C

Answer 3

∫1/（x²-1)
=1/2∫1/(x-1)-1/(x+1)dx
=1/2(ln|x-1|-ln|x+1|)+C
=1/2ln|(x-1)/(x+1)|+C

Answer 4

1/x²+x+1定积分
∫(1/x²+x+1)
=-1/x+x²/2+x+c