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1、§4．1不定積分的概念與性質(zhì)§4?3分部積分法設(shè)函數(shù)u?u(x)及v?v(x)具有連續(xù)導(dǎo)數(shù)?那么?兩個函數(shù)乘積的導(dǎo)數(shù)公式為(uv)??u?v?uv??移項得uv??(uv)??u?v?對這個等式兩邊求不定積分?得?uv?dx?uv??u?vdx?或?udv?uv??vdu?這個公式稱為分部積分公式?分部積分過程:?uv?dx??udv?uv??vdu?uv??u?vdx?????例1?xcosxdx??xdsinx?xsinx??sinxdx?xsinx?cosx?C?例2?xexdx??xdex?xex??exdx?xex?ex?C?例3?x2exdx??x2dex?x2ex?
2、?exdx2?x2ex?2?xexdx?x2ex?2?xdex?x2ex?2xex?2?exdx?x2ex?2xex?2ex?C?ex(x2?2x?2)?C?1111例4?xlnxdx??lnxdx2?x2lnx??x2?dx222x1111?x2lnx??xdx?x2lnx?x2?C?2224例5?arccosxdx?xarccosx??xdarccosx1?xarccosx??xdx1?x21?1?xarccosx??(1?x2)2d(1?x2)?xarccosx?1?x2?C?21111例6?xarctanxdx??arctanxdx2?x2arctanx??x2?dx22
3、21?x2111?x2arctanx??(1?)dx221?x2111?x2arctanx?x?arctanx?C?222例7求?exsinxdx?1§4．1不定積分的概念與性質(zhì)解因為?exsinxdx??sinxdex?exsinx??exdsinx?exsinx??excosxdx?exsinx??cosxdex?exsinx?excosx??exdcosx?exsinx?excosx??exdcosx?exsinx?excosx??exsinxdx?1所以?exsinxdx?ex(sinx?cosx)?C?2例8求?sec3xdx?解因為?sec3xdx??secx?sec
4、2xdx??secxdtanx?secxtanx??secxtan2xdx?secxtanx??secx(sec2x?1)dx?secxtanx??sec3xdx??secxdx?secxtanx?ln
5、secx?tanx
6、??sec3xdx?1所以?sec3xdx?(secxtanx?ln
7、secx?tanx
8、)?C?2dx例9求I???其中n為正整數(shù)?n(x2?a2)ndx1x解I???arctan?C?1x2?a2aa當n?1時，用分部積分法?有dxxx2???2(n?1)?dx(x2?a2)n?1(x2?a2)n?1(x2?a2)nx1a2??2(n?1)?[?]dx?(
9、x2?a2)n?1(x2?a2)n?1(x2?a2)nx即I??2(n?1)(I?a2I)?n?1(x2?a2)n?1n?1n2§4．1不定積分的概念與性質(zhì)1x于是I?[?(2n?3)I]?n2a2(n?1)(x2?a2)n?1n?11x以此作為遞推公式?并由I?arctan?C即可得I?1aan例10求?exdx?解令x?t2?則?dx?2tdt?于?exdx?2?tetdt?2et(t?1)?C?2ex(x?1)?C??exdx??exd(x)2?2?xexdx?2?xdex?2xex?2?exdx?2xex?2ex?C?2ex(x?1)?C?第一換元法與分部積分法的比較:共
10、同點是第一步都是湊微分令?(x)?u?f[?(x)]??(x)dx??f[?(x)]d?(x)?f(u)du??u(x)v?(x)dx??u(x)dv(x)?u(x)v(x)??v(x)du(x)?哪些積分可以用分部積分法？?xcosxdx??xexdx??x2exdx??xlnxdx??arccosxdx??xarctanxdx??exsinxdx??sec3xdx??2xex2dx??ex2dx2??eudu??????x2exdx??x2dex?x2ex??exdx2?????3
