证明题,关于反三角函数及其导函数的,求证步骤完美的话有加分^.^已知,f(x)=arctan(x)函数满足,（x^2+1）*f'(x)=1求证(x^2+1)f^(k+1)(x)+2kxf^(k)(x)+k(k-1)f^(k-1)(x)=0k=1,2......k不等于0

　　用数学归纳法可证.
当n = 1时,对(x^2+1)*f'(x)=1求导,得
(x^2+1)*f"(x)+2x f'(x) = 0,
命题成立.
　　设n = k 时,命题成立,即
(x^2+1)f^(k+1)(x)+2kxf^(k)(x)+k(k-1)f^(k-1)(x) = 0,
则n = k+1 时,对上式再求一次导数,得
　…………
(x^2+1)f^(k+2)(x)+2(k+1)xf^(k+1)(x)+(k+1)kf^(k)(x) = 0,
依归纳法原理,命题得证.