[高等数学]分部积分法

一、知识点

设函数 u = u ( x ) u=u(x) u=u(x) 及 v = v ( x ) v=v(x) v=v(x) 具有连续函数。那么，两个函数乘积的导数公式为 ( u v ) ′ = u ′ v + u v ′ ， (uv)'=u'v+uv'， (uv)′=u′v+uv′， 移项得 u v ′ = ( u v ) ′ − u ′ v . uv'=(uv)'-u'v. uv′=(uv)′−u′v.

对上述等式两边求不定积分，得 ∫ u v ′ d x = u v − ∫ u ′ v d x . (1) \int uv'dx=uv-\int u'vdx.\tag{1} ∫uv′dx=uv−∫u′vdx.(1)

公式 ( 1 ) (1) (1) 称为分部积分公式.

为简便起见，公式 ( 1 ) (1) (1) 可以写成以下形式： ∫ u d v = u v − ∫ v d u . (2) \int udv=uv-\int vdu.\tag{2} ∫udv=uv−∫vdu.(2)

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二、练习题（求不定积分）

( 1 ) ∫ x s i n x d x = − ∫ x d c o s x = − ( x c o s x − ∫ c o s x d x ) = s i n x − x c o s x + C \begin{aligned} (1) &\int xsinxdx\\ &=-\int xdcosx\\ &=-(xcosx-\int cosxdx)\\ &=sinx-xcosx+C \end{aligned} (1)​∫xsinxdx=−∫xdcosx=−(xcosx−∫cosxdx)=sinx−xcosx+C​

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( 2 ) ∫ l n x d x = x l n x − ∫ x d ( l n x ) = x l n x − ∫ d x = x l n x − x + C \begin{aligned} (2) &\int lnxdx\\ &=xlnx-\int xd(lnx)\\ &=xlnx-\int dx\\ &=xlnx-x+C \end{aligned} (2)​∫lnxdx=xlnx−∫xd(lnx)=xlnx−∫dx=xlnx−x+C​

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( 3 ) ∫ a r c s i n x d x = x a r c s i n x − ∫ x d ( a r c s i n x ) = x a r c s i n x − ∫ x 1 − x 2 d x = x a r c s i n x + 1 2 ∫ d ( 1 − x 2 ) 1 − x 2 = x a r c s i n x + 1 − x 2 + C \begin{aligned} (3) &\int arcsinxdx\\ &=xarcsinx-\int xd(arcsinx)\\ &=xarcsinx-\int \frac{x}{\sqrt{1-x^2}}dx\\ &=xarcsinx+\frac{1}{2}\int \frac{d(1-x^2)}{\sqrt{1-x^2}}\\ &=xarcsinx+\sqrt{1-x^2}+C \end{aligned} (3)​∫arcsinxdx=xarcsinx−∫xd(arcsinx)=xarcsinx−∫1−x2 ​x​dx=xarcsinx+21​∫1−x2 ​d(1−x2)​=xarcsinx+1−x2 ​+C​

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( 4 ) ∫ x e − x d x = − x e − x − ∫ e − x d x = − x e − x − e − x + C \begin{aligned} (4) &\int xe^{-x}dx\\ &=-xe^{-x}-\int e^{-x}dx\\ &=-xe^{-x}-e^{-x}+C \end{aligned} (4)​∫xe−xdx=−xe−x−∫e−xdx=−xe−x−e−x+C​

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( 5 ) ∫ x 2 l n x d x = ∫ 1 3 l n x d ( x 3 ) = x 3 l n x 3 − 1 3 ∫ x 3 d ( l n x ) = x 3 l n x 3 − 1 3 ∫ x 2 d x = x 3 l n x 3 − x 3 9 + C \begin{aligned} (5) &\int x^2lnxdx\\ &=\int \frac{1}{3}lnxd(x^3)\\ &=\frac{x^3lnx}{3}-\frac{1}{3}\int x^3d(lnx)\\ &=\frac{x^3lnx}{3}-\frac{1}{3}\int x^2dx\\ &=\frac{x^3lnx}{3}-\frac{x^3}{9}+C \end{aligned} (5)​∫x2lnxdx=∫31​lnxd(x3)=3x3lnx​−31​∫x3d(lnx)=3x3lnx​−31​∫x2dx=3x3lnx​−9x3​+C​

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( 6 ) ∫ e − x c o s x d x = ∫ e − x d s i n x = e − x s i n x − ∫ s i n x d ( e − x ) = e − x s i n x + ∫ e − x s i n x d x = e − x s i n x − ∫ e − x d c o s x = e − x s i n x − e − x c o s x − ∫ c o s x d ( e − x ) ∴ ∫ e − x c o s x d x = e − x s i n x − e − x c o s x 2 + C \begin{aligned} (6) &\int e^{-x}cosxdx\\ &=\int e^{-x}dsinx\\ &=e^{-x}sinx-\int sinxd(e^{-x})\\ &=e^{-x}sinx+\int e^{-x}sinxdx\\ &=e^{-x}sinx-\int e^{-x}dcosx\\ &=e^{-x}sinx-e^{-x}cosx-\int cosx d(e^{-x})\\ &\therefore \int e^{-x}cosxdx=\frac{e^{-x}sinx-e^{-x}cosx}{2}+C \end{aligned} (6)​∫e−xcosxdx=∫e−xdsinx=e−xsinx−∫sinxd(e−x)=e−xsinx+∫e−xsinxdx=e−xsinx−∫e−xdcosx=e−xsinx−e−xcosx−∫cosxd(e−x)∴∫e−xcosxdx=2e−xsinx−e−xcosx​+C​

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( 7 ) ∫ e − 2 x s i n x 2 d x = − 2 ∫ e − 2 x d c o s x 2 = − 2 ( e − 2 x c o s x 2 − ∫ c o s x 2 d e − 2 x ) = − 2 ( e − 2 x c o s x 2 + 2 ∫ e − 2 x c o s x 2 d x ) = − 2 e − 2 x c o s x 2 − 4 ∫ e − 2 x 2 d ( s i n x 2 ) = − 2 e − 2 x c o s x 2 − 8 ( s i n x 2 − ∫ s i n x 2 d e − 2 x ) = − 2 e − 2 x c o s x 2 − 8 e − 2 x s i n x 2 − 16 ∫ e − 2 x s i n x 2 ∴ ∫ e − 2 x s i n x 2 d x = − 2 17 e − 2 x ( c o s x 2 + 4 s i n x 2 ) + C \begin{aligned} (7) &\int e^{-2x}sin\frac{x}{2}dx\\ &=-2\int e^{-2x}dcos\frac{x}{2}\\ &=-2(e^{-2x}cos\frac{x}{2}-\int cos\frac{x}{2}de^{-2x})\\ &=-2(e^{-2x}cos\frac{x}{2}+2\int e^{-2x} cos\frac{x}{2}dx)\\ &=-2e^{-2x}cos\frac{x}{2}-4\int e^{-2x}2d(sin\frac{x}{2})\\ &=-2e^{-2x}cos\frac{x}{2}-8(sin\frac{x}{2}-\int sin\frac{x}{2}de^{-2x})\\ &=-2e^{-2x}cos\frac{x}{2}-8e^{-2x}sin\frac{x}{2}-16\int e^{-2x}sin\frac{x}{2}\\ &\therefore \int e^{-2x}sin\frac{x}{2}dx=-\frac{2}{17}e^{-2x}(cos\frac{x}{2}+4sin\frac{x}{2})+C \end{aligned} (7)​∫e−2xsin2x​dx=−2∫e−2xdcos2x​=−2(e−2xcos2x​−∫cos2x​de−2x)=−2(e−2xcos2x​+2∫e−2xcos2x​dx)=−2e−2xcos2x​−4∫e−2x2d(sin2x​)=−2e−2xcos2x​−8(sin2x​−∫sin2x​de−2x)=−2e−2xcos2x​−8e−2xsin2x​−16∫e−2xsin2x​∴∫e−2xsin2x​dx=−172​e−2x(cos2x​+4sin2x​)+C​

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( 8 ) ∫ x c o s x 2 d x = 2 ∫ x d ( s i n x 2 ) = 2 x s i n x 2 − 2 ∫ s i n x 2 d x = 2 x s i n x 2 + 4 c o s x 2 + C \begin{aligned} (8) &\int xcos\frac{x}{2}dx\\ &=2\int xd(sin\frac{x}{2})\\ &=2xsin\frac{x}{2}-2\int sin\frac{x}{2}dx\\ &=2xsin\frac{x}{2}+4cos\frac{x}{2}+C \end{aligned} (8)​∫xcos2x​dx=2∫xd(sin2x​)=2xsin2x​−2∫sin2x​dx=2xsin2x​+4cos2x​+C​

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( 9 ) ∫ x 2 a r c t a n x d x = 1 3 ∫ a r c t a n x d ( x 3 ) = 1 3 x 3 a r c t a n x − 1 3 ∫ x 3 d ( a r c t a n x ) = 1 3 x 3 a r c t a n x − 1 3 ∫ x 3 + x − x 1 + x 2 d x = 1 3 x 3 a r c t a n x − 1 3 ∫ x d x + 1 3 ∫ x d x x 2 + 1 = 1 3 x 3 a r c t a n x − x 2 6 + 1 6 ∫ d ( x 2 + 1 ) x 2 + 1 = 1 3 x 3 a r c t a n x − x 2 6 + l n ( x 2 + 1 ) 6 + C \begin{aligned} (9) &\int x^2arctanxdx\\ &=\frac{1}{3}\int arctanxd(x^3)\\ &=\frac{1}{3}x^3arctanx-\frac{1}{3}\int x^3d(arctanx)\\ &=\frac{1}{3}x^3arctanx-\frac{1}{3}\int \frac{x^3+x-x}{1+x^2}dx\\ &=\frac{1}{3}x^3arctanx-\frac{1}{3}\int xdx+\frac{1}{3}\int \frac{xdx}{x^2+1}\\ &=\frac{1}{3}x^3arctanx-\frac{x^2}{6}+\frac{1}{6}\int \frac{d(x^2+1)}{x^2+1}\\ &=\frac{1}{3}x^3arctanx-\frac{x^2}{6}+\frac{ln(x^2+1)}{6}+C \end{aligned} (9)​∫x2arctanxdx=31​∫arctanxd(x3)=31​x3arctanx−31​∫x3d(arctanx)=31​x3arctanx−31​∫1+x2x3+x−x​dx=31​x3arctanx−31​∫xdx+31​∫x2+1xdx​=31​x3arctanx−6x2​+61​∫x2+1d(x2+1)​=31​x3arctanx−6x2​+6ln(x2+1)​+C​

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( 10 ) ∫ x t a n 2 x d x = ∫ x ( s e c 2 x − 1 ) d x = ∫ x s e c 2 x d x − ∫ x d x = ∫ x d ( t a n x ) − ∫ x d x = x t a n x − ∫ t a n x d x − 1 2 x 2 = x t a n x + l n ∣ c o s x ∣ − 1 2 x 2 + C \begin{aligned} (10) &\int xtan^2xdx\\ &=\int x(sec^2x-1)dx\\ &=\int xsec^2xdx-\int xdx\\ &=\int xd(tanx) -\int xdx\\ &=xtanx-\int tanxdx-\frac{1}{2}x^2\\ &=xtanx+ln|cosx|-\frac{1}{2}x^2+C \end{aligned} (10)​∫xtan2xdx=∫x(sec2x−1)dx=∫xsec2xdx−∫xdx=∫xd(tanx)−∫xdx=xtanx−∫tanxdx−21​x2=xtanx+ln∣cosx∣−21​x2+C​

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( 11 ) ∫ x 2 c o s x d x = ∫ x 2 d ( s i n x ) = x 2 s i n x − ∫ s i n x d x 2 = x 2 s i n x − 2 ∫ x s i n x d x = x 2 s i n x − 2 ∫ x d ( c o s x ) = x 2 s i n x − 2 ( x c o s x − ∫ c o s x d x ) = x 2 s i n x + 2 x s i n x − 2 s i n x + C \begin{aligned} (11) &\int x^2cosxdx\\ &=\int x^2d(sinx)\\ &=x^2sinx-\int sinxdx^2\\ &=x^2sinx-2\int xsinxdx\\ &=x^2sinx-2\int xd(cosx)\\ &=x^2sinx-2(xcosx-\int cosxdx)\\ &=x^2sinx+2xsinx-2sinx+C \end{aligned} (11)​∫x2cosxdx=∫x2d(sinx)=x2sinx−∫sinxdx2=x2sinx−2∫xsinxdx=x2sinx−2∫xd(cosx)=x2sinx−2(xcosx−∫cosxdx)=x2sinx+2xsinx−2sinx+C​

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( 12 ) ∫ t e − 2 t d x = − 1 2 ∫ t d ( e − 2 t ) = − 1 2 ( t e − 2 t − ∫ e − 2 t d t ) = − 1 2 t e − 2 t − 1 4 e − 2 t + C \begin{aligned} (12) &\int te^{-2t}dx\\ &=-\frac{1}{2}\int td(e^{-2t})\\ &=-\frac{1}{2}(te^{-2t}-\int e^{-2t}dt)\\ &=-\frac{1}{2}te^{-2t}-\frac{1}{4}e^{-2t}+C \end{aligned} (12)​∫te−2tdx=−21​∫td(e−2t)=−21​(te−2t−∫e−2tdt)=−21​te−2t−41​e−2t+C​

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( 13 ) ∫ l n 2 x d x = x ( l n x ) 2 − ∫ x d [ ( l n x ) 2 ] = x ( l n x ) 2 − 2 ∫ l n x d x = x ( l n x ) 2 − 2 [ x l n x − ∫ x d ( l n x ) ] = x ( l n x ) 2 − 2 x l n x − 2 x + C \begin{aligned} (13) &\int ln^2xdx\\ &=x(lnx)^2-\int xd[(lnx)^2]\\ &=x(lnx)^2-2\int lnxdx\\ &=x(lnx)^2-2[xlnx-\int xd(lnx)]\\ &=x(lnx)^2-2xlnx-2x+C \end{aligned} (13)​∫ln2xdx=x(lnx)2−∫xd[(lnx)2]=x(lnx)2−2∫lnxdx=x(lnx)2−2[xlnx−∫xd(lnx)]=x(lnx)2−2xlnx−2x+C​

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( 14 ) ∫ x s i n x c o s x d x = 1 2 ∫ x s i n 2 x d x = − 1 4 ∫ x d ( c o s 2 x ) = − 1 4 x c o s 2 x + 1 4 ∫ c o s 2 x d x = 1 8 s i n 2 x − 1 4 x c o s 2 x + C \begin{aligned} (14) &\int xsinxcosxdx\\ &=\frac{1}{2}\int xsin2xdx\\ &=-\frac{1}{4}\int xd(cos2x)\\ &=-\frac{1}{4}xcos2x+\frac{1}{4}\int cos2xdx\\ &=\frac{1}{8}sin2x-\frac{1}{4}xcos2x+C \end{aligned} (14)​∫xsinxcosxdx=21​∫xsin2xdx=−41​∫xd(cos2x)=−41​xcos2x+41​∫cos2xdx=81​sin2x−41​xcos2x+C​

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( 15 ) ∫ x 2 c o s 2 x 2 d x = 1 2 ∫ x 2 ( c o s x + 1 ) d x = 1 2 ∫ x 2 c o s x d x + 1 2 ∫ x 2 d x = x 3 6 + 1 2 ∫ x 2 d ( s i n x ) = x 3 6 + 1 2 [ x 2 s i n x − ∫ s i n x d ( x 2 ) ] = x 3 6 + x 2 s i n x 2 − ∫ x s i n x d x = x 3 6 + x 2 s i n x 2 + ∫ x d ( c o s x ) = x 3 6 + x 2 s i n x 2 + x c o s x − ∫ c o s x d x = x 3 6 + x 2 s i n x 2 + x c o s x − s i n x + C \begin{aligned} (15) &\int x^2cos^2\frac{x}{2}dx\\ &=\frac{1}{2}\int x^2(cosx+1)dx\\ &=\frac{1}{2}\int x^2cosxdx+\frac{1}{2}\int x^2dx\\ &=\frac{x^3}{6}+\frac{1}{2}\int x^2d(sinx)\\ &=\frac{x^3}{6}+\frac{1}{2}[x^2sinx-\int sinxd(x^2)]\\ &=\frac{x^3}{6}+\frac{x^2sinx}{2}-\int xsinxdx\\ &=\frac{x^3}{6}+\frac{x^2sinx}{2}+\int xd(cosx)\\ &=\frac{x^3}{6}+\frac{x^2sinx}{2}+xcosx-\int cosxdx\\ &=\frac{x^3}{6}+\frac{x^2sinx}{2}+xcosx-sinx+C \end{aligned} (15)​∫x2cos22x​dx=21​∫x2(cosx+1)dx=21​∫x2cosxdx+21​∫x2dx=6x3​+21​∫x2d(sinx)=6x3​+21​[x2sinx−∫sinxd(x2)]=6x3​+2x2sinx​−∫xsinxdx=6x3​+2x2sinx​+∫xd(cosx)=6x3​+2x2sinx​+xcosx−∫cosxdx=6x3​+2x2sinx​+xcosx−sinx+C​

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( 16 ) ∫ x l n ( x − 1 ) d x = 1 2 ∫ l n ( x − 1 ) d ( x 2 ) = 1 2 x 2 l n ( x − 1 ) − 1 2 ∫ x 2 d [ l n ( x − 1 ) ] = 1 2 x 2 l n ( x − 1 ) − 1 2 ∫ x 2 x − 1 d x = 1 2 x 2 l n ( x − 1 ) − 1 2 ∫ x 2 − 1 + 1 x − 1 d x = 1 2 x 2 l n ( x − 1 ) − 1 2 ∫ ( x + 1 ) d x − 1 2 ∫ d x x − 1 = 1 2 x 2 l n ( x − 1 ) − 1 4 ( x + 1 ) 2 − l n ∣ x − 1 ∣ 2 + C \begin{aligned} (16) &\int xln(x-1)dx\\ &=\frac{1}{2}\int ln(x-1)d(x^2)\\ &=\frac{1}{2}x^2ln(x-1)-\frac{1}{2}\int x^2d[ln(x-1)]\\ &=\frac{1}{2}x^2ln(x-1)-\frac{1}{2}\int \frac{x^2}{x-1}dx\\ &=\frac{1}{2}x^2ln(x-1)-\frac{1}{2}\int \frac{x^2-1+1}{x-1}dx\\ &=\frac{1}{2}x^2ln(x-1)-\frac{1}{2}\int (x+1)dx-\frac{1}{2}\int \frac{dx}{x-1}\\ &=\frac{1}{2}x^2ln(x-1)-\frac{1}{4}(x+1)^2-\frac{ln|x-1|}{2}+C \end{aligned} (16)​∫xln(x−1)dx=21​∫ln(x−1)d(x2)=21​x2ln(x−1)−21​∫x2d[ln(x−1)]=21​x2ln(x−1)−21​∫x−1x2​dx=21​x2ln(x−1)−21​∫x−1x2−1+1​dx=21​x2ln(x−1)−21​∫(x+1)dx−21​∫x−1dx​=21​x2ln(x−1)−41​(x+1)2−2ln∣x−1∣​+C​

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( 17 ) ∫ ( x 2 − 1 ) s i n 2 x d x = ∫ x 2 s i n 2 x d x − ∫ s i n 2 x d x = − 1 2 ∫ x 2 d ( c o s 2 x ) + 1 2 c o s 2 x = 1 2 c o s 2 x − 1 2 x 2 c o s 2 x + 1 2 ∫ c o s 2 x d ( x 2 ) = 1 2 c o s 2 x − 1 2 x 2 c o s 2 x + ∫ x c o s 2 x d x = 1 2 c o s 2 x − 1 2 x 2 c o s 2 x + 1 2 ∫ x d ( s i n 2 x ) = 1 2 c o s 2 x − 1 2 x 2 c o s 2 x + 1 2 x s i n 2 x − 1 2 ∫ s i n 2 x d x = 1 2 c o s 2 x − 1 2 x 2 c o s 2 x + 1 2 x s i n 2 x + 1 4 c o s 2 x + C = ( 3 4 − 1 2 x 2 ) c o s 2 x + 1 2 x s i n 2 x + C \begin{aligned} (17) &\int (x^2-1)sin2xdx\\ &=\int x^2sin2xdx-\int sin2xdx\\ &=-\frac{1}{2}\int x^2d(cos2x)+\frac{1}{2}cos2x\\ &=\frac{1}{2}cos2x-\frac{1}{2}x^2cos2x+\frac{1}{2}\int cos2xd(x^2)\\ &=\frac{1}{2}cos2x-\frac{1}{2}x^2cos2x+\int xcos2xdx\\ &=\frac{1}{2}cos2x-\frac{1}{2}x^2cos2x+\frac{1}{2}\int xd(sin2x)\\ &=\frac{1}{2}cos2x-\frac{1}{2}x^2cos2x+\frac{1}{2}xsin2x-\frac{1}{2}\int sin2xdx\\ &=\frac{1}{2}cos2x-\frac{1}{2}x^2cos2x+\frac{1}{2}xsin2x+\frac{1}{4}cos2x+C\\ &=(\frac{3}{4}-\frac{1}{2}x^2)cos2x+\frac{1}{2}xsin2x+C \end{aligned} (17)​∫(x2−1)sin2xdx=∫x2sin2xdx−∫sin2xdx=−21​∫x2d(cos2x)+21​cos2x=21​cos2x−21​x2cos2x+21​∫cos2xd(x2)=21​cos2x−21​x2cos2x+∫xcos2xdx=21​cos2x−21​x2cos2x+21​∫xd(sin2x)=21​cos2x−21​x2cos2x+21​xsin2x−21​∫sin2xdx=21​cos2x−21​x2cos2x+21​xsin2x+41​cos2x+C=(43​−21​x2)cos2x+21​xsin2x+C​

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( 18 ) ∫ l n 3 x x 2 d x = − ∫ ( l n x ) 3 d ( 1 x ) = − ( l n x ) 3 x + ∫ 1 x d [ ( l n x ) 3 ] = − ( l n x ) 3 x + 3 ∫ ( l n x ) 2 x 2 d x = − ( l n x ) 3 x − 3 ∫ ( l n x ) 2 d ( 1 x ) = − ( l n x ) 3 x − 3 ( l n x ) 3 x + 3 ∫ d [ ( l n x ) 2 ] x = − ( l n x ) 3 x − 3 ( l n x ) 3 x + 6 ∫ l n x x 2 d x = − ( l n x ) 3 x − 3 ( l n x ) 3 x − 6 ∫ l n x d ( 1 x ) = − ( l n x ) 3 x − 3 ( l n x ) 3 x − 6 l n x x + 6 ∫ d ( l n x ) x = − 1 x [ 6 l n x + 3 ( l n x ) 2 + ( l n x ) 3 − 6 ] + C \begin{aligned} (18) &\int \frac{ln^3x}{x^2}dx\\ &=-\int (lnx)^3d(\frac{1}{x})\\ &=-\frac{(lnx)^3}{x}+\int \frac{1}{x}d[(lnx)^3]\\ &=-\frac{(lnx)^3}{x}+3\int \frac{(lnx)^2}{x^2}dx\\ &=-\frac{(lnx)^3}{x}-3\int (lnx)^2d(\frac{1}{x})\\ &=-\frac{(lnx)^3}{x}-\frac{3(lnx)^3}{x}+3\int \frac{d[(lnx)^2]}{x}\\ &=-\frac{(lnx)^3}{x}-\frac{3(lnx)^3}{x}+6\int \frac{lnx}{x^2}dx\\ &=-\frac{(lnx)^3}{x}-\frac{3(lnx)^3}{x}-6\int lnxd(\frac{1}{x})\\ &=-\frac{(lnx)^3}{x}-\frac{3(lnx)^3}{x}-\frac{6lnx}{x}+6\int \frac{d(lnx)}{x}\\ &=-\frac{1}{x}[6lnx+3(lnx)^2+(lnx)^3-6]+C \end{aligned} (18)​∫x2ln3x​dx=−∫(lnx)3d(x1​)=−x(lnx)3​+∫x1​d[(lnx)3]=−x(lnx)3​+3∫x2(lnx)2​dx=−x(lnx)3​−3∫(lnx)2d(x1​)=−x(lnx)3​−x3(lnx)3​+3∫xd[(lnx)2]​=−x(lnx)3​−x3(lnx)3​+6∫x2lnx​dx=−x(lnx)3​−x3(lnx)3​−6∫lnxd(x1​)=−x(lnx)3​−x3(lnx)3​−x6lnx​+6∫xd(lnx)​=−x1​[6lnx+3(lnx)2+(lnx)3−6]+C​

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( 19 ) ∫ e 3 x d x ( 令 t = 3 3 ) = ∫ e t d t 3 = 3 ∫ e t t 2 d t = 3 ∫ t 2 d e t = 3 t 2 e t − 3 ∫ e t d t 2 = 3 t 2 e t − 6 ∫ t e t d t = 3 t 2 e t − 6 ∫ t d e t = 3 t 2 e t − 6 t e t + 6 ∫ e t d t = 3 x 2 3 e x 3 − 2 x 3 e x 3 + 2 e x 3 + C \begin{aligned} (19) &\int e^{3\sqrt{x}}dx\\ &(令 t=\sqrt[3]{3})\\ &=\int e^tdt^3\\ &=3\int e^tt^2dt\\ &=3\int t^2de^t\\ &=3t^2e^t-3\int e^tdt^2\\ &=3t^2e^t-6\int te^tdt\\ &=3t^2e^t-6\int tde^t\\ &=3t^2e^t-6te^t+6\int e^tdt\\ &=3x^{\frac{2}{3}}e^{\sqrt[3]{x}}-2\sqrt[3]{x}e^{\sqrt[3]{x}}+2e^{\sqrt[3]{x}}+C \end{aligned} (19)​∫e3x ​dx(令t=33 ​)=∫etdt3=3∫ett2dt=3∫t2det=3t2et−3∫etdt2=3t2et−6∫tetdt=3t2et−6∫tdet=3t2et−6tet+6∫etdt=3x32​e3x ​−23x ​e3x ​+2e3x ​+C​

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( 20 ) ∫ c o s l n x d x = x c o s ( l n x ) − ∫ x d [ c o s ( l n x ) ] = x c o s ( l n x ) + ∫ s i n ( l n x ) d x = x c o s ( l n x ) + x s i n ( l n x ) − ∫ x d [ s i n ( l n x ) ] = x c o s ( l n x ) + x s i n ( l n x ) − ∫ c o s ( l n x ) d x ∴ ∫ c o s ( l n x ) d x = x c o s ( l n x ) + x s i n ( l n x ) 2 + C \begin{aligned} (20) &\int coslnxdx\\ &=xcos(lnx)-\int xd[cos(lnx)]\\ &=xcos(lnx)+\int sin(lnx)dx\\ &=xcos(lnx)+xsin(lnx)-\int xd[sin(lnx)]\\ &=xcos(lnx)+xsin(lnx)-\int cos(lnx)dx\\ &\therefore \int cos(lnx)dx=\frac{xcos(lnx)+xsin(lnx)}{2}+C \end{aligned} (20)​∫coslnxdx=xcos(lnx)−∫xd[cos(lnx)]=xcos(lnx)+∫sin(lnx)dx=xcos(lnx)+xsin(lnx)−∫xd[sin(lnx)]=xcos(lnx)+xsin(lnx)−∫cos(lnx)dx∴∫cos(lnx)dx=2xcos(lnx)+xsin(lnx)​+C​

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学习资料：《高等数学（第六版）》 ，同济大学数学系 编

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