Question

求arctanx/（1+x^2)^(3/2)的不定积分，急急急！！

Answer 1

∫ 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

Answer 2

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