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求不定积分∫[(1+1/x^2)(1-x^2)+xsinx]dx
人气:239 ℃ 时间:2020-05-24 17:19:46
解答
∫[(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
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