求不定积分∫[（1+1/x^2)(1-x^2)+xsinx]dx

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/28 10:47:09

∫[（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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