求arctanx/(1+x^2)^(3/2)的不定积分,急!

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 00:46:03
求arctanx/(1+x^2)^(3/2)的不定积分,急!
xJ@_eDI2.L1Bld1bdakA[EQjqSw)4|o Hu9swg;JEszdeJǶ,bA(5RfCgd*kT4_}DʲsȠMEqMbeaGWsd*Jy,4A/V/,LQ"xLK46gi rh"0'"|Ɍ<>!%L̊^x*6Q{L,^X +X농OToFqF~|f7.܂r]X.{2nFl?l!]

求arctanx/(1+x^2)^(3/2)的不定积分,急!
求arctanx/(1+x^2)^(3/2)的不定积分,急!

求arctanx/(1+x^2)^(3/2)的不定积分,急!
∫ arctanx / (1+x²)^(3/2) dx
= ∫ arctanx d[x/√(x²+1)],分部积分法,∫ dx/(1+x²)^(3/2) = x/√(x²+1)
= [x/√(x²+1)]arctanx - ∫ x/√(x²+1) d(arctanx),(arcanx)' = 1/(x²+1)
= x*arctanx / √(x²+1) - ∫ x/(x²+1)^(3/2) dx
= x*arctanx / √(x²+1) - (1/2)∫ d(x²+1)/(x²+1)^(3/2)
= x*arctanx / √(x²+1) - (1/2)*(x²+1)^(-3/2+1) / (-3/2+1) + C
= x*arctanx / √(x²+1) - (1/2)(-2)(x²+1)^(-1/2) + C
= x*arctanx / √(x²+1) + 1/√(x²+1) + C
= (x*arctanx + 1) / √(x² + 1) + C

用换元法吧
设t=arctanx
则dx=sec^2tdt
则原式=tcostdt
这个不定积分用分部积分很容易求的
最后回代就行了