ch04第四章不定積分-03第三節(jié)分部積分法

第三節(jié)部積分布圖

★分部積分公 ★幾點說★★★★★★★★★★★★★★★★★★例★★例★例★例★★習(xí)題-★內(nèi)容要

udvuvuvdxuv

分部積分法實質(zhì)上就是求兩函數(shù)乘積的導(dǎo)數(shù)(或微分)的逆運算.一般地,下列類型的被積函數(shù)?？紤]應(yīng)用分部積分法(m,n都是正整數(shù)).xnsinmxxnarcsin

xncosmxenxcosmxxn(lnx)xn例題選1E01)求不定積分xcosxdxx2x解一令ucosx,xdx 2 x2 xcosxdxcosxd22顯然

解二令ux,cosxdxdsinxxcosxdxxdsinxxsinxsinxdxxsinxcosx2E02)求不定積分x2exdx解ux2exdxdexx2exdxx2dexx2ex2xexdxx2ex2xdexx2ex2(xexex)注：若被積函數(shù)是冪函數(shù)(指數(shù)為正整數(shù))與指數(shù)函數(shù)或正(余)弦函數(shù)的乘積,可設(shè)冪函數(shù)為u,而將其余部分湊微分進入微分號,使得應(yīng)用分部積分公式后,冪函數(shù)的冪次降低一3E03)求不定積分xarctanxdxx2x解令uarctanx,xdx 2 x2

xarctanxdxarctanxd2

2arctanx

d(arctanx)2

2arctanx

21x2

1 2arctanx211x2dx

arctanx (xarctanx) 4E04)求不定積分x3lnxdx x4解令ulnx,xdx 4 x4 1

1 xlnxdxlnxd44

lnx4xdx4

lnx

注：若被積函數(shù)是冪函數(shù)與對數(shù)函數(shù)或反三角函數(shù)的乘積可設(shè)對數(shù)函數(shù)或反三角函數(shù)為u,而將冪函數(shù)湊微分進入微分號,使得應(yīng)用分部積分公式后,對數(shù)函數(shù)或反三角函數(shù)消例5 求不定積分exsinxdx解exsindxsindexexsinxexd(sinx)exsinxexcosexsinxcosxdexexsinx(excosxexdcosex(sinxcosx)exsinexsindxex(sinxcosx)2注(余)弦函數(shù)的乘積,udv可隨意選取,但在兩次分部積分中,u,以便經(jīng)過兩次分部積分后產(chǎn)生循環(huán)式,從而解出所求積分.例6 求不定積分sin(lnx)dx解sin(lnx)dxxsin(lnxxd[sin(lnxsin(lnx)xcos(lnx)1xsin(lnx)dxx[sin(lnx)cos(lnx)]2 例7 求不定積分sec3xdx解sec3xdxsecxdtanxsecxtanxsecxtan2由于上式右端的第三項就是所求的積分sec3xdx2，便得sec3xdx1(secxtanxln|secxtanx|)C.218求不定積分arcsinx111xx解arcsinxdx11xxxx

1x

1xd

1x

1xx1x1

1x

xx1x9求不定積分xxx1x1解xarctanxdxarctan 111111 arctanx1

1x2d1 arctanx1

11111 arctanx1111tan2 dxxtant sec2tdtsectdtln(secttant)11tan2

1x2)原式

11

1x2)例10 求不定積分exdx解令txxt2dx2tdt于exdx2ettdt2tdet2tet2et2tet2etC2et(t1)C2ex

xx例 求不定積分

x)dx解令t

x,xtx)dxln(1t)dt2t2ln(1t)t2dln(1t)t2ln(1t)

t1 t2ln(1t)(t1)dt x1x

tln(1t) tln(1t) 2

x)

xC.312I3

ex1/

3解法 先分部積分，后換元.設(shè)uex1/3,3

1dxdu1x2/3ex1/3dx,v3x2/3, 于是I3x2/3ex131ex13 xt3dx3t2dt,ex1/3dx3t2etdt3t2et6tetdt3t2et6tetetdt3(t22t2)et代入上式,I3x2/3ex1/33

2)ex1/3C

1)ex1/3 xx解法 先換元,后分部積分.設(shè)xt3,dx3t2dt,xxeteI再設(shè)utdvetdt,

2dt3tetx2xI3tet3etdt3tet3etc3(3例13 求不定積分(1x)arcsin(1x)dx.解令t1x,則dxx2x

1)ex1/31t原式tarcsintdt1t

1t21t1t arcsint1t1t

1t21t arcsintt1t2x 2x其中CC1例14 求不定積分In

(x2a2

,n為正整數(shù)解用分部積分法，當n1(x2a2

(x2a2

2x2(n1)(x2a2)nx

(x2a2

1)(x2a2

(x2a2)n In1

(x2a2

2(n

a2In于 I 2a2(n

(x

x

(2n(2n

1arctanxC即可得I 例15 已知f(x)的一個原函數(shù)是ex2,求xf(x)dx解xf(x)dxxdf(x)xf(xf根據(jù)題意f(x)dxex2C再注意到f(x)dxfxf(x)2xex2xf(x)dxxf(x)f(x)dx2x2ex2ex216求不定積分

sinxxcos3xsincos2

解先折成兩個不定積分，再利用分部積分法原式esinxxcosxdx

sinxdxxdesinxesinxd1cos2

cosxesinxesinxdxesincos17求不定積分sinxln(tan

esinxdxxesinx1esinxcos解sinxln(tanx)dxln(tanx)dcosxcosxln(tanxcosxdln(tancosxln(tanx)1dxcosxln(tanx)ln|cscxcotx|sin例 求不定積分(x2)2dx解選ux2ex于x2ex

2x

1

2x

1

2(x2)2dxxedx2xex2x2d(xe x2ex

x2exx xx2exxexdxx2exx xx2exxexexdxx2exxexexx2

x注:本題選ux

比選u(x2)2更能使解題方便19計算不定積分xln解lnxx放在右列,列表如下()lnx()11x xlnxdxlnx1x211x2dx1x2lnx1xdx1x2lnx1x2 x 20計算不定積分ln解lnx可看作乘積形式1lnx將lnx放在左列,1放在右列，列表如下()lnx()1xxlnxdxxlnx1xdxxlnxxx21計算不定積分xsin解x和sinx都是易求原函數(shù)的函數(shù),x放左列,sinx放右列列表如下()xsin cos()0sinxsinxdxxcosx1(sinx)cxcosxsinx22計算不定積分excosxdx解函數(shù)exco