y = ∫ 0 x t f ( t 2 − x 2 ) d t y = \int _0^x t f( t^2 - x^2)dt y=∫0xtf(t2−x2)dt
设 t 2 − x 2 = u , 那么 t = u + x 2 , d t = d u 2 u + x 2 ， 并且当 t = x 时 u = 0 , 当 t = 0 时， u = − x 2 设 t^2 - x^2 = u,那么t =\sqrt{u+x^2},dt =\frac{du}{2\sqrt{u+x^2}}，\\ 并且当t=x时u=0,当t=0时，u= -x^2 设t2−x2=u,那么t=u+x2,dt=2u+x2du，并且当t=x时u=0,当t=0时，u=−x2
y = ∫ 0 x t f ( t 2 − x 2 ) d t = ∫ 0 − x 2 u + x 2 f ( u ) d u 2 u + x 2 = 1 2 ∫ − x 2 0 f ( u ) d u = − 1 2 ∫ 0 − x 2 f ( u ) d u y = \int _0^x t f( t^2 - x^2)dt = \int^{-x^2}_0 \sqrt{u+x^2}f(u) \frac{du}{2\sqrt{u+x^2}} =\frac{1}{2} \int_{-x^2}^0 f(u)du =\\ -\frac{1}{2} \int^{-x^2}_0 f(u)du y=∫0xtf(t2−x2)dt=∫0−x2u+x2f(u)2u+x2du=21∫−x20f(u)du=−21∫0−x2f(u)du
d ( − 1 2 ∫ 0 − x 2 f ( u ) d u ) d x = x f ( − x 2 ) \frac{d(-\frac{1}{2} \int^{-x^2}_0 f(u)du)}{dx} = xf(-x^2) dxd(−21∫0−x2f(u)du)=xf(−x2)
 A ~ C)