不定积分∫[1/2cosx+2/(1+x2)-(1/3)x]dx的答案

网友回答

∫[1/(2 cos x) + 2/(1 + x²) - x/3] dx
= ∫ (1/2)(sec x) + 2[1/(1 + x²)] - x/3 dx
= (1/2)∫ sec x + 4[1/(1 + x²)] - 2x/3 dx
把他们分出来（因为各有各的解决方法）,
∫ sec x dx,
乘于 (sec x + tan x)/(sec x + tan x),
= ∫ sec x[(sec x + tan x)/(sec x + tan x)] dx
= ∫ (sec² x + tan x sec x)/(sec x + tan x) dx
让 u = sec x + tan x
du = tan x sec x + sec² x dx
所以,∫ (sec² x + tan x sec x)/(sec x + tan x) dx
= ∫ 1/u du
= ln u= ln (sec x + tan x)
∫4[1/(1 + x²)] dx
让 x = tan θ => θ = arc tan x
dx = sec² θ dθ
所以,∫4[1/(1 + x²)] dx
= 4∫ [1/(1 + tan² θ)](sec² θ) dθ
= 4∫ (sec² θ)/(sec² θ) dθ
= 4∫ dθ= 4θ= 4 arc tan x
∫ -2x/3 dx
= -x²/3
把所有东西合起来,
(1/2)∫ sec x + 4[1/(1 + x²)] - 2x/3 dx
= (1/2)[ln (sec x + tan x) + (4 arc tan x) - (x²/3) + C],C为任何值.
= ln √(sec x + tan x) + (2 arc tan x) - (x²/6) + D ,D为任何值.