∫[（1+1/x^2)(1-x^2)+xsinx]dx
=∫(1/x^2-x^2 + xsinx) dx
= -1/x - 1/3*x^3 +∫xsinx dx
= -1/x - 1/3*x^3 -∫xdcosx
= -1/x - 1/3*x^3 -xcosx + ∫cosx dx
= -1/x - 1/3*x^3 -xcosx + sinx + C
求不定积分∫[（1+1/x^2)(1-x^2)+xsinx]dx
求不定积分∫[（1+1/x^2)(1-x^2)+xsinx]dx
数学人气：797 ℃时间：2020-05-24 17:19:46
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