若∫f(x)dx=x^2+C则∫xf(1-x^2)dx=若∫f(x)dx=x^2+C则∫xf(1-x^2)dx=A2(1-x^2)^2+CB-2(1-x^2)^2+CC1/2*(1-x^2)^2+CD-1/2*(1-x^2)^2+C

Question

若∫f ( x )dx=x^2+C 则∫x f ( 1-x^2)dx=
若∫f ( x )dx=x^2+C 则∫x f ( 1-x^2)dx= A 2(1-x^2)^2+C B -2(1-x^2)^2+C C 1/2 * (1-x^2)^2+C D - 1/2 * (1-x^2)^2+C

Answer 1

答：
∫xf(1-x^2) dx
=∫-1/2*f(1-x^2) d(1-x^2)
=-1/2*(1-x^2)^2 + C
即选D.