已知(1-x^2)f"(x)+xf'(x)=x/(x+1)

用分部积分配方法：.
(1-x^2)f "(x) + xf '(x) ＝ x/(x+1) 方程两边乘以(1-x^2)^(-3/2)
==> (1-x^2)^(-1/2) f "(x) + x(1-x^2)^(-3/2) f '(x) ＝ x(1-x^2)^(-3/2)/(x+1)
==> [ (1-x^2)^(-1/2) f '(x) ] ' ＝ x(1-x^2)^(-3/2)/(x+1)
==> (1-x^2)^(-1/2) f '(x) ＝ ∫ x(1-x^2)^(-3/2)/(x+1) dx
==> (1-x^2)^(-1/2) f '(x) ＝ ∫ (x-x^2)(1-x^2)^(-5/2) dx
==> (1-x^2)^(-1/2) f '(x) ＝ (1-x^3)(1-x^2)^(-3/2) / 3 + C1
==> f '(x) ＝ (1-x^3)(1-x^2)^(-1) / 3 + C1*(1-x^2)^(1/2)
==> f (x) ＝ ∫ [ (1-x^3)(1-x^2)^(-1) / 3 + C1*(1-x^2)^(1/2) ] dx
==> f (x) ＝ ∫ [ (1+x+x^2)/(3(1+x)) + C1*(1-x^2)^(1/2) ] dx
==> f (x) ＝ x^2/6 + ln(1+x) /3 + C1*[ x(1-x^2)^(1/2)/2 + arcsin(x)/2 ] + C2
==> f (x) ＝ x^2/6 + ln(1+x) /3 + C1*[ x(1-x^2)^(1/2) + arcsin(x) ] + C2 （C1为常数,1/2可去掉)
==> f (x) ＝ x^2 / 6 + ln(1+x) /3 + C1*[ x√(1-x^2) + arcsin(x) ] + C2 （到此步已验证正确)
根据初始条件解得:C2＝0,C1＝?
f(0.5)给的有点问题,C1的结果很复杂,不再计算.

设p(x)=f'(x) 则f''(x)=p'(x)
(1-x^2)p'(x)+xp(x)=x/(x+1)
(1-x^2)p'(x)=-xp(x)
p'/p=-x/(1-x^2)
lnp=1/2ln|1-x^2|+lnC
p=C(1-x^2)^(1/2)
p'=C'(1-x^2)^(1/2)-2xC/2(1-x^2)^(-1/2)
(1-x...