是用第一类换元法
∫xf(1-x^2)dx＝－1/2×∫f(1-x^2)×(1-x^2)'dx＝－1/2×∫f(1-x^2)d(1-x^2),令t＝1-x^2,则
∫xf(1-x^2)dx＝－1/2×∫f(t)dt＝－1/2×(t^2＋C1)＝－1/2×(1-x^2)^2＋C
C＝1/2×C1