高等数学（第七版）同济大学习题4-3 个人解答

高等数学（第七版）同济大学习题4-3

求下列不定积分：\begin{aligned}&\ 求下列不定积分：&\end{aligned} 求下列不定积分：

(1)∫xsinxdx； (2)∫lnxdx；(3)∫arcsinxdx； (4)∫xe−xdx；(5)∫x2lnxdx； (6)∫e−xcosxdx；(7)∫e−2xsinx2dx； (8)∫xcosx2dx；(9)∫x2arctanxdx； (10)∫xtan2xdx；(11)∫x2cosxdx； (12)∫te−2tdt；(13)∫ln2xdx； (14)∫xsinxcosxdx；(15)∫x2cos2x2dx； (16)∫xln(x−1)dx；(17)∫(x2−1)sin2xdx； (18)∫ln3xx2dx；(19)∫ex3dx； (20)∫coslnxdx；(21)∫(arcsinx)2dx； (22)∫exsin2xdx；(23)∫xln2xdx； (24)∫e3x+9dx；\begin{aligned} &\ \ (1)\ \ \int xsin\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \int ln\ xdx；\\\\ &\ \ (3)\ \ \int arcsin\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ \int xe^{-x}dx；\\\\ &\ \ (5)\ \ \int x^2ln\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ \int e^{-x}cos\ xdx；\\\\ &\ \ (7)\ \ \int e^{-2x}sin\ \frac{x}{2}dx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ \int xcos\ \frac{x}{2}dx；\\\\ &\ \ (9)\ \ \int x^2arctan\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)\ \ \int xtan^2\ xdx；\\\\ &\ \ (11)\ \ \int x^2cos\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (12)\ \ \int te^{-2t}dt；\\\\ &\ \ (13)\ \ \int ln^2\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (14)\ \ \int xsin\ xcos\ xdx；\\\\\ &\ \ (15)\ \ \int x^2cos^2\ \frac{x}{2}dx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (16)\ \ \int xln(x-1)dx；\\\\ &\ \ (17)\ \ \int (x^2-1)sin\ 2xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (18)\ \ \int \frac{ln^3\ x}{x^2}dx；\\\\ &\ \ (19)\ \ \int e^{\sqrt[3]{x}}dx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (20)\ \ \int cos\ ln\ xdx；\\\\ &\ \ (21)\ \ \int (arcsin\ x)^2dx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (22)\ \ \int e^xsin^2\ xdx；\\\\ &\ \ (23)\ \ \int xln^2\ xdx；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (24)\ \ \int e^{\sqrt{3x+9}}dx；\\\\ & \end{aligned} (1) ∫xsin xdx； (2) ∫ln xdx； (3) ∫arcsin xdx； (4) ∫xe−xdx； (5) ∫x2ln xdx； (6) ∫e−xcos xdx； (7) ∫e−2xsin 2xdx； (8) ∫xcos 2xdx； (9) ∫x2arctan xdx； (10) ∫xtan2 xdx； (11) ∫x2cos xdx； (12) ∫te−2tdt； (13) ∫ln2 xdx； (14) ∫xsin xcos xdx； (15) ∫x2cos2 2xdx； (16) ∫xln(x−1)dx； (17) ∫(x2−1)sin 2xdx； (18) ∫x2ln3 xdx； (19) ∫e3xdx； (20) ∫cos ln xdx； (21) ∫(arcsin x)2dx； (22) ∫exsin2 xdx； (23) ∫xln2 xdx； (24) ∫e3x+9dx；

解：

(1)设u=x，dv=sinxdx，则du=dx，v=−cosx，得∫xsinxdx=−xcos⁡x+∫cosxdx=−xcosx+sinx+C(2)设u=lnx，dv=dx，则du=1xdx，v=x，得∫lnxdx=xlnx−∫dx=xln⁡x−x+C(3)设u=arcsinx，dv=dx，则du=11−x2dx，v=x，得∫arcsinxdx=xarcsinx−∫x1−x2dx=xarcsinx+1−x2+C(4)设u=x，dv=e−xdx，则du=dx，v=−e−x，得∫xe−xdx=−xe−x+∫e−xdx=−e−x(x+1)+C(5)设u=lnx，dv=x2dx，则du=1xdx，v=13x3，得∫x2lnxdx=13x3lnx−13∫x2dx=13x3lnx−19x3+C(6)设u=cosx，dv=e−xdx，则du=−sinxdx，v=−e−x，得∫e−xcosxdx=−e−xcosx−∫e−xsinxdx，求∫e−xsinxdx，设u=sinx，dv=e−xdx，则du=cosxdx，v=−e−x，得∫e−xsinxdx=−e−xsinx+∫e−xcosxdx，代入原式，得∫e−xcosxdx=−e−xcosx−∫e−xsinxdx=−e−xcosx+e−xsinx−∫e−xcosxdx，则∫e−xcosxdx=12e−x(sinx−cosx)+C(7)设u=sinx2，dv=e−2xdx，则du=12cosx2dx，v=−12e−2x，得∫e−2xsinx2dx=−12e−2xsinx2+14∫e−2xcosx2dx，求∫e−2xcosx2dx，设u=cosx2，dv=e−2xdx，则du=−12sinx2，v=−12e−2x，得∫e−2xcosx2dx=−12e−2xcosx2−14∫e−2xsinx2dx，代入原式，得∫e−2xsinx2dx=−12e−2xsinx2+14∫e−2xcosx2dx=−12e−2xsinx2−18e−2xcosx2−116∫e−2xsinx2dx，则∫e−2xsinx2dx=−817e−2xsinx2−217e−2xcosx2dx+C(8)设u=x，dv=cosx2dx，则du=dx，v=2sinx2，得∫xcosx2dx=2xsinx2−2∫sinx2dx=2xsinx2+4cosx2+C(9)设u=arctanx，dv=x2dx，则du=11+x2dx，v=13x3，得∫x2arctanxdx=13x3arctanx−13∫x31+x2dx，求∫x31+x2dx，令u=1+x2，则x=u−1，dx=12u−1du，得∫x31+x2dx=12∫u−1udu=12∫du−12∫1udu=12u−12ln∣u∣+C=12(1+x2)−12ln(1+x2)+C，代入原式，得∫x2arctanxdx=13x3arctanx−13∫x31+x2dx=13x3arctanx−16(1+x2)+16ln(1+x2)+C=13x3arctanx+16ln(1+x2)−16x2+C(10)∫xtan2xdx=∫x(sec2x−1)dx=∫xsec2xdx−∫xdx，令u=x，dv=sec2xdx，则du=dx，v=tanx，得∫xsec2xdx−∫xdx=xtanx−∫tanxdx−12x2，求∫tanxdx，∫tanxdx=−∫1cosxd(cosx)=−ln∣cosx∣+C，代入原式，得xtanx+ln∣cosx∣−12x2+C(11)设u=x2，dv=cosxdx，则du=2xdx，v=sinx，得∫x2cosxdx=x2sinx−2∫xsinxdx，求∫xsinxdx，设u=x，dv=sinxdx，则du=dx，v=−cosx，得∫xsinxdx=−xcosx+∫cosxdx=−xcosx+sinx+C，代入原式，得∫x2cosxdx=x2sinx−2∫xsinxdx=x2sinx+2xcosx−2sinx+C=(x2−2)sinx+2xcosx+C(12)设u=t，dv=e−2tdt，则du=dt，v=−12e−2t，得∫te−2tdt=−12te−2t+12∫e−2tdt=−12te−2t−14e−2t+C=−12e−2t(t+12)+C(13)设u=ln2x，dv=dx，则du=2lnxxdx，v=x，得∫ln2xdx=xln2x−2∫lnxdx，求∫lnxdx，设u=lnx，x=eu，dx=eudu，得∫lnxdx=∫ueudu=xlnx−x+C，代入原式，得∫ln2xdx=xln2x−2∫lnxdx=xln2x−2xlnx+2x+C(14)设u=xsinx，dv=cosxdx，则du=(sinx+xcosx)dx，v=sinx，得∫xsinxcosxdx=xsin2x−∫(sin2x+xsinxcosx)dx=xsin2x−∫sin2xdx−∫xsinxcosxdx，求∫sin2xdx=−14sin2x+12x+C，代入原式，得∫xsinxcosxdx=xsin2x+14sin2x−12x−∫xsinxcosxdx，则∫xsinxcosxdx=12xsin2x+18sin2x−14x+C=−14xcos2x+18sin2x+C(15)设u=cos2x2，dv=x2dx，则du=−12sinxdx，v=13x3，得∫x2cos2x2dx=13x3cos2x2+16∫x3sinxdx（1−1）求∫x3sinxdx，设u=x3，dv=sinxdx，则du=3x2dx，v=−cosx，得∫x3sinxdx=−x3cosx+3∫x2cosxdx（1−2）求∫x2cosxdx，设u=x2，dv=cosxdx，则du=2xdx，v=sinx，得∫x2cosxdx=x2sinx−2∫xsinxdx（1−3）求∫xsinxdx，设u=x，dv=sinxdx，则du=dx，v=−cosx，得∫xsinxdx=−xcosx+∫cosxdx=−xcosx+sinx+C，代入（1−3）式，得∫x2cosxdx=x2sinx−2∫xsinxdx=x2sinx+2xcosx−2sinx+C，代入（1−2）式，得∫x3sinxdx=−x3cosx+3∫x2cosxdx=−x3cosx+3x2sinx+6xcosx−6sinx+C，代入（1−1）式，得∫x2cos2x2dx=13x3cos2x2+16∫x3sinxdx=13x3cos2x2−16x3cosx+12x2sinx+xcosx−sinx+C=16x3+12x2sinx+xcosx−sinx+C(16)设u=ln(x−1)，dv=xdx，则du=1x−1dx，v=12x2，得∫xln(x−1)dx=12x2ln(x−1)−12∫x2x−1dx，求∫x2x−1dx，设u=x−1，则dx=du，得∫x2x−1dx=∫(u+1)2udu=∫(u+2+1u)du=12u2+2u+ln∣u∣+C=12(x+1)2+ln(x−1)+C，代入原式，∫xln(x−1)dx=12x2ln(x−1)−12∫x2x−1dx=12x2ln(x−1)−14(x+1)2−12ln(x−1)+C=12(x2−1)ln(x−1)−14x2−12x+C(17)设u=x2−1，dv=sin2xdx，则du=2xdx，v=−12cos2x，得∫(x2−1)sin2xdx=−12x2cos2x+12cos2x+∫xcos2xdx，求∫xcos2xdx设u=x，dv=cos2xdx，则du=dx，v=12sin2x，得∫xcos2xdx=12xsin2x−12∫sin2xdx=12xsin2x+14cos2x+C，代入原式，得∫(x2−1)sin2xdx=−12x2cos2x+12cos2x+∫xcos2xdx=−12x2cos2x+34cos2x+12xsin2x+C(18)设u=ln3x，dv=1x2dx，则du=3ln2xxdx，v=−1x，得∫ln3xx2dx=−ln3xx+3∫ln2xx2dx（1−1），求∫ln2xx2dx设u=ln2x，dv=1x2dx，则du=2lnxx，v=−1x，得∫ln2xx2dx=−ln2xx+2∫lnxx2dx（1−2），求∫lnxx2dx设u=lnx，dv=1x2dx，则du=1xdx，v=−1x，得∫lnxx2dx=−lnxx+∫1x2dx=−lnxx−1x+C，代入（1−2）式，得∫ln2xx2dx=−ln2xx+2∫lnxx2dx=−ln2xx−2lnxx−2x+C，代入（1−1）式，得∫ln3xx2dx=−ln3xx+3∫ln2xx2dx=−ln3xx−3ln2xx−6lnxx−6x+C(19)令u=x3，x=u3，则dx=3u2du，得∫ex3dx=3∫u2eudu，设t=u2，dv=eudu，则dt=2udu，v=eu，得3∫u2eudu=3u2eu−6∫ueudu，设t=u，dv=eudu，则dt=du，v=eu，得3∫u2eudu=3u2eu−6∫ueudu=3u2eu−6ueu+6eu+C=3ex3(x23−2x3+2)+C(20)设u=lnx，x=eu，则dx=eudu，得∫coslnxdx=∫eucosudu，设t=cosu，dv=eudu，则dt=−sinudu，v=eu，得∫eucosudu=eucosu+∫eusinudu=eucosu+eusinu−∫eucosudu，∫eucosudu=12eu(sinu+cosu)+C=12x(sinlnx+coslnx)+C(21)设u=arcsinx，x=sinu，dx=cosudu，得∫(arcsinx)2dx=∫u2cosudu，设t=u2，dv=cosudu，则dt=2udu，v=sinu，得∫u2cosudu=u2sinu−2∫usinudu，设t=u，dv=sinudu，则dt=du，v=−cosu，得∫u2cosudu=u2sinu−2∫usinudu=u2sinu+2ucosu−2∫cosudu)=u2sinu+2ucosu−2sinu+C得∫(arcsinx)2dx=x(arcsinx)2+21−x2arcsinx−2x+C(22)设u=sin2x，dv=exdx，则du=sin2xdx，v=ex，得∫exsin2xdx=exsin2x−∫exsin2xdx（1−1）求∫exsin2xdx，设u=sin2x，dv=exdx，则du=2cos2xdx，v=ex，得∫exsin2xdx=exsin2x−2∫excos2xdx（1−2），求∫excos2xdx，设u=cos2x，，dv=exdx，则du=−2sin2xdx，v=ex，得∫excos2xdx=excos2x+2∫exsin2xdx，代入（1−2）式，得∫exsin2xdx=15exsin2x−25excos2x+C，代入（1−1）式，得∫exsin2xdx=exsin2x−∫exsin2xdx=exsin2x−15exsin2x+25excos2x+C=12ex−15exsin2x−110excos2x+C(23)设u=lnx，则x=eu，dx=eudu，得∫xln2xdx=∫u2e2udu，设t=u2，dv=e2udu，则dt=2udu，v=12e2u，得∫u2e2udu=12u2e2u−∫ue2udu，求∫ue2udu，设t=u，dv=e2udu，则dt=du，v=12e2u，得∫ue2udu=12ue2u−12∫e2udu=12ue2u−14e2u+C，代入原式，得∫u2e2udu=12u2e2u−∫ue2udu=12u2e2u−12ue2u+14e2u+C，∫xln2xdx=12x2ln2x−12x2lnx+14x2+C=12x2(ln2x−lnx+12)+C(24)设u=3x+9，则x=13u2−3，dx=23udu，得∫e3x+9dx=23∫ueudu，设t=u，dv=eudu，则dt=du，v=eu，得23∫ueudu=23(ueu−∫eudu)=23eu(u−1)+C=23e3x+9(3x+9−1)+C\begin{aligned} &\ \ (1)\ 设u=x，dv=sin\ xdx，则du=dx，v=-cos\ x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int xsin\ xdx=-x\cos\ x+\int cos\ xdx=-xcos\ x+sin\ x+C\\\\ &\ \ (2)\ 设u=ln\ x，dv=dx，则du=\frac{1}{x}dx，v=x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int ln\ xdx=xln\ x-\int dx=x\ln\ x-x+C\\\\ &\ \ (3)\ 设u=arcsin\ x，dv=dx，则du=\frac{1}{\sqrt{1-x^2}}dx，v=x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int arcsin\ xdx=xarcsin\ x-\int \frac{x}{\sqrt{1-x^2}}dx=xarcsin\ x+\sqrt{1-x^2}+C\\\\ &\ \ (4)\ 设u=x，dv=e^{-x}dx，则du=dx，v=-e^{-x}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx=-e^{-x}(x+1)+C\\\\ &\ \ (5)\ 设u=ln\ x，dv=x^2dx，则du=\frac{1}{x}dx，v=\frac{1}{3}x^3，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2ln\ xdx=\frac{1}{3}x^3ln\ x-\frac{1}{3}\int x^2dx=\frac{1}{3}x^3ln\ x-\frac{1}{9}x^3+C\\\\ &\ \ (6)\ 设u=cos\ x，dv=e^{-x}dx，则du=-sin\ xdx，v=-e^{-x}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int e^{-x}cos\ xdx=-e^{-x}cos\ x-\int e^{-x}sin\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 求\int e^{-x}sin\ xdx，设u=sin\ x，dv=e^{-x}dx，则du=cos\ xdx，v=-e^{-x}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int e^{-x}sin\ xdx=-e^{-x}sin\ x+\int e^{-x}cos\ xdx，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int e^{-x}cos\ xdx=-e^{-x}cos\ x-\int e^{-x}sin\ xdx=-e^{-x}cos\ x+e^{-x}sin\ x-\int e^{-x}cos\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 则\int e^{-x}cos\ xdx=\frac{1}{2}e^{-x}(sin\ x-cos\ x)+C\\\\ &\ \ (7)\ 设u=sin\ \frac{x}{2}，dv=e^{-2x}dx，则du=\frac{1}{2}cos\ \frac{x}{2}dx，v=-\frac{1}{2}e^{-2x}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int e^{-2x}sin\ \frac{x}{2}dx=-\frac{1}{2}e^{-2x}sin\ \frac{x}{2}+\frac{1}{4}\int e^{-2x}cos\ \frac{x}{2}dx，\\\\ &\ \ \ \ \ \ \ \ \ 求\int e^{-2x}cos\ \frac{x}{2}dx，设u=cos\ \frac{x}{2}，dv=e^{-2x}dx，则du=-\frac{1}{2}sin\ \frac{x}{2}，v=-\frac{1}{2}e^{-2x}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int e^{-2x}cos\ \frac{x}{2}dx=-\frac{1}{2}e^{-2x}cos\ \frac{x}{2}-\frac{1}{4}\int e^{-2x}sin\ \frac{x}{2}dx，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int e^{-2x}sin\ \frac{x}{2}dx=-\frac{1}{2}e^{-2x}sin\ \frac{x}{2}+\frac{1}{4}\int e^{-2x}cos\ \frac{x}{2}dx=-\frac{1}{2}e^{-2x}sin\ \frac{x}{2}-\frac{1}{8}e^{-2x}cos\ \frac{x}{2}-\frac{1}{16}\int e^{-2x}sin\ \frac{x}{2}dx，\\\\ &\ \ \ \ \ \ \ \ \ 则\int e^{-2x}sin\ \frac{x}{2}dx=-\frac{8}{17}e^{-2x}sin\ \frac{x}{2}-\frac{2}{17}e^{-2x}cos\ \frac{x}{2}dx+C\\\\ &\ \ (8)\ 设u=x，dv=cos\ \frac{x}{2}dx，则du=dx，v=2sin\ \frac{x}{2}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int xcos\ \frac{x}{2}dx=2xsin\ \frac{x}{2}-2\int sin\ \frac{x}{2}dx=2xsin\ \frac{x}{2}+4cos\ \frac{x}{2}+C\\\\ &\ \ (9)\ 设u=arctan\ x，dv=x^2dx，则du=\frac{1}{1+x^2}dx，v=\frac{1}{3}x^3，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2arctan\ xdx=\frac{1}{3}x^3arctan\ x-\frac{1}{3}\int \frac{x^3}{1+x^2}dx，\\\\ &\ \ \ \ \ \ \ \ \ 求\int \frac{x^3}{1+x^2}dx，令u=1+x^2，则x=\sqrt{u-1}，dx=\frac{1}{2\sqrt{u-1}}du，\\\\ &\ \ \ \ \ \ \ \ \ 得\int \frac{x^3}{1+x^2}dx=\frac{1}{2}\int \frac{u-1}{u}du=\frac{1}{2}\int du-\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}u-\frac{1}{2}ln\ |u|+C=\frac{1}{2}(1+x^2)-\frac{1}{2}ln(1+x^2)+C，\\\\ &\ \ \ \ \ \ \ \ \ 代入原式，得\int x^2arctan\ xdx=\frac{1}{3}x^3arctan\ x-\frac{1}{3}\int \frac{x^3}{1+x^2}dx=\frac{1}{3}x^3arctan\ x-\frac{1}{6}(1+x^2)+\frac{1}{6}ln(1+x^2)+C\\\\ &\ \ \ \ \ \ \ \ \ =\frac{1}{3}x^3arctan\ x+\frac{1}{6}ln(1+x^2)-\frac{1}{6}x^2+C\\\\ &\ \ (10)\ \int xtan^2\ xdx=\int x(sec^2\ x-1)dx=\int xsec^2\ xdx-\int xdx，令u=x，dv=sec^2\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 则du=dx，v=tan\ x，得\int xsec^2\ xdx-\int xdx=xtan\ x-\int tan\ xdx-\frac{1}{2}x^2，求\int tan\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ \int tan\ xdx=-\int \frac{1}{cos\ x}d(cos\ x)=-ln\ |cos\ x|+C，代入原式，得\\\\ &\ \ \ \ \ \ \ \ \ xtan\ x+ln\ |cos\ x|-\frac{1}{2}x^2+C\\\\ &\ \ (11)\ 设u=x^2，dv=cos\ xdx，则du=2xdx，v=sin\ x，得\int x^2cos\ xdx=x^2sin\ x-2\int xsin\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 求\int xsin\ xdx，设u=x，dv=sin\ xdx，则du=dx，v=-cos\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \int xsin\ xdx=-xcos\ x+\int cos\ xdx=-xcos\ x+sin\ x+C，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2cos\ xdx=x^2sin\ x-2\int xsin\ xdx=x^2sin\ x+2xcos\ x-2sin\ x+C=(x^2-2)sin\ x+2xcos\ x+C\\\\ &\ \ (12)\ 设u=t，dv=e^{-2t}dt，则du=dt，v=-\frac{1}{2}e^{-2t}，\\\\ &\ \ \ \ \ \ \ \ \ 得\int te^{-2t}dt=-\frac{1}{2}te^{-2t}+\frac{1}{2}\int e^{-2t}dt=-\frac{1}{2}te^{-2t}-\frac{1}{4}e^{-2t}+C=-\frac{1}{2}e^{-2t}\left(t+\frac{1}{2}\right)+C\\\\ &\ \ (13)\ 设u=ln^2\ x，dv=dx，则du=\frac{2ln\ x}{x}dx，v=x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int ln^2\ xdx=xln^2\ x-2\int ln\ xdx，求\int ln\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 设u=ln\ x，x=e^u，dx=e^udu，得\int ln\ xdx=\int ue^udu=xln\ x-x+C，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int ln^2\ xdx=xln^2\ x-2\int ln\ xdx=xln^2\ x-2xln\ x+2x+C\\\\ &\ \ (14)\ 设u=xsin\ x，dv=cos\ xdx，则du=(sin\ x+xcos\ x)dx，v=sin\ x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int xsin\ xcos\ xdx=xsin^2\ x-\int (sin^2\ x+xsin\ xcos\ x)dx=xsin^2\ x-\int sin^2\ xdx-\int xsin\ xcos\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 求\int sin^2\ xdx=-\frac{1}{4}sin\ 2x+\frac{1}{2}x+C，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int xsin\ xcos\ xdx=xsin^2\ x+\frac{1}{4}sin\ 2x-\frac{1}{2}x-\int xsin\ xcos\ xdx，\\\\ &\ \ \ \ \ \ \ \ \ 则\int xsin\ xcos\ xdx=\frac{1}{2}xsin^2\ x+\frac{1}{8}sin\ 2x-\frac{1}{4}x+C=-\frac{1}{4}xcos\ 2x+\frac{1}{8}sin\ 2x+C\\\\ &\ \ (15)\ 设u=cos^2\ \frac{x}{2}，dv=x^2dx，则du=-\frac{1}{2}sin\ xdx，v=\frac{1}{3}x^3，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2cos^2\ \frac{x}{2}dx=\frac{1}{3}x^3cos^2\ \frac{x}{2}+\frac{1}{6}\int x^3sin\ xdx（1-1）\\\\ &\ \ \ \ \ \ \ \ \ 求\int x^3sin\ xdx，设u=x^3，dv=sin\ xdx，则du=3x^2dx，v=-cos\ x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^3sin\ xdx=-x^3cos\ x+3\int x^2cos\ xdx（1-2）\\\\ &\ \ \ \ \ \ \ \ \ 求\int x^2cos\ xdx，设u=x^2，dv=cos\ xdx，则du=2xdx，v=sin\ x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2cos\ xdx=x^2sin\ x-2\int xsin\ xdx（1-3）\\\\ &\ \ \ \ \ \ \ \ \ 求\int xsin\ xdx，设u=x，dv=sin\ xdx，则du=dx，v=-cos\ x，\\\\ &\ \ \ \ \ \ \ \ \ 得\int xsin\ xdx=-xcos\ x+\int cos\ xdx=-xcos\ x+sin\ x+C，代入（1-3）式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2cos\ xdx=x^2sin\ x-2\int xsin\ xdx=x^2sin\ x+2xcos\ x-2sin\ x+C，代入（1-2）式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^3sin\ xdx=-x^3cos\ x+3\int x^2cos\ xdx=-x^3cos\ x+3x^2sin\ x+6xcos\ x-6sin\ x+C，\\\\ &\ \ \ \ \ \ \ \ \ 代入（1-1）式，\\\\ &\ \ \ \ \ \ \ \ \ 得\int x^2cos^2\ \frac{x}{2}dx=\frac{1}{3}x^3cos^2\ \frac{x}{2}+\frac{1}{6}\int x^3sin\ xdx=\frac{1}{3}x^3cos^2\ \frac{x}{2}-\frac{1}{6}x^3cos\ x+\frac{1}{2}x^2sin\ x+xcos\ x-sin\ x+C=\\\\ &\ \ \ \ \ \ \ \ \ \frac{1}{6}x^3+\frac{1}{2}x^2sin\ x+xcos\ x-sin\ x+C\\\\ &\ \ (16)\ 设u=ln(x-1)，dv=xdx，则du=\frac{1}{x-1}dx，v=\frac{1}{2}x^2，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int xln(x-1)dx=\frac{1}{2}x^2ln(x-1)-\frac{1}{2}\int \frac{x^2}{x-1}dx，求\int \frac{x^2}{x-1}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=x-1，则dx=du，得\int \frac{x^2}{x-1}dx=\int \frac{(u+1)^2}{u}du=\int \left(u+2+\frac{1}{u}\right)du=\frac{1}{2}u^2+2u+ln\ |u|+C=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}(x+1)^2+ln(x-1)+C，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ \ \int xln(x-1)dx=\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}{4}(x+1)^2-\frac{1}{2}ln(x-1)+C=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}(x^2-1)ln(x-1)-\frac{1}{4}x^2-\frac{1}{2}x+C\\\\ &\ \ (17)\ 设u=x^2-1，dv=sin\ 2xdx，则du=2xdx，v=-\frac{1}{2}cos\ 2x，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int (x^2-1)sin\ 2xdx=-\frac{1}{2}x^2cos\ 2x+\frac{1}{2}cos\ 2x+\int xcos\ 2xdx，求\int xcos\ 2xdx\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=x，dv=cos\ 2xdx，则du=dx，v=\frac{1}{2}sin\ 2x，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int xcos\ 2xdx=\frac{1}{2}xsin\ 2x-\frac{1}{2}\int sin\ 2xdx=\frac{1}{2}xsin\ 2x+\frac{1}{4}cos\ 2x+C，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int (x^2-1)sin\ 2xdx=-\frac{1}{2}x^2cos\ 2x+\frac{1}{2}cos\ 2x+\int xcos\ 2xdx=-\frac{1}{2}x^2cos\ 2x+\frac{3}{4}cos\ 2x+\frac{1}{2}xsin\ 2x+C\\\\ &\ \ (18)\ 设u=ln^3\ x，dv=\frac{1}{x^2}dx，则du=\frac{3ln^2\ x}{x}dx，v=-\frac{1}{x}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int \frac{ln^3\ x}{x^2}dx=-\frac{ln^3\ x}{x}+3\int \frac{ln^2\ x}{x^2}dx（1-1），求\int \frac{ln^2\ x}{x^2}dx\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=ln^2\ x，dv=\frac{1}{x^2}dx，则du=\frac{2ln\ x}{x}，v=-\frac{1}{x}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int \frac{ln^2\ x}{x^2}dx=-\frac{ln^2\ x}{x}+2\int \frac{ln\ x}{x^2}dx（1-2），求\int \frac{ln\ x}{x^2}dx\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=ln\ x，dv=\frac{1}{x^2}dx，则du=\frac{1}{x}dx，v=-\frac{1}{x}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int \frac{ln\ x}{x^2}dx=-\frac{ln\ x}{x}+\int \frac{1}{x^2}dx=-\frac{ln\ x}{x}-\frac{1}{x}+C，代入（1-2）式，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int \frac{ln^2\ x}{x^2}dx=-\frac{ln^2\ x}{x}+2\int \frac{ln\ x}{x^2}dx=-\frac{ln^2\ x}{x}-\frac{2ln\ x}{x}-\frac{2}{x}+C，代入（1-1）式，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int \frac{ln^3\ x}{x^2}dx=-\frac{ln^3\ x}{x}+3\int \frac{ln^2\ x}{x^2}dx=-\frac{ln^3\ x}{x}-\frac{3ln^2\ x}{x}-\frac{6ln\ x}{x}-\frac{6}{x}+C\\\\ &\ \ (19)\ 令u=\sqrt[3]{x}，x=u^3，则dx=3u^2du，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^{\sqrt[3]{x}}dx=3\int u^2e^udu，设t=u^2，dv=e^udu，则dt=2udu，v=e^u，\\\\ &\ \ \ \ \ \ \ \ \ \ 得3\int u^2e^udu=3u^2e^u-6\int ue^udu，设t=u，dv=e^udu，则dt=du，v=e^u，\\\\ &\ \ \ \ \ \ \ \ \ \ 得3\int u^2e^udu=3u^2e^u-6\int ue^udu=3u^2e^u-6ue^u+6e^u+C=3e^{\sqrt[3]{x}}(x^{\frac{2}{3}}-2\sqrt[3]{x}+2)+C\\\\ &\ \ (20)\ 设u=ln\ x，x=e^u，则dx=e^udu，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int cos\ ln\ xdx=\int e^ucos\ udu，设t=cos\ u，dv=e^udu，则dt=-sin\ udu，v=e^u，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^ucos\ udu=e^ucos\ u+\int e^usin\ udu=e^ucos\ u+e^usin\ u-\int e^ucos\ udu，\\\\ &\ \ \ \ \ \ \ \ \ \ \int e^ucos\ udu=\frac{1}{2}e^u(sin\ u+cos\ u)+C=\frac{1}{2}x(sin\ ln\ x+cos\ ln\ x)+C\\\\ &\ \ (21)\ 设u=arcsin\ x，x=sin\ u，dx=cos\ udu，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int (arcsin\ x)^2dx=\int u^2cos\ udu，设t=u^2，dv=cos\ udu，则dt=2udu，v=sin\ u，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int u^2cos\ udu=u^2sin\ u-2\int usin\ udu，设t=u，dv=sin\ udu，则dt=du，v=-cos\ u，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int u^2cos\ udu=u^2sin\ u-2\int usin\ udu=u^2sin\ u+2ucos\ u-2\int cos\ udu)=\\\\ &\ \ \ \ \ \ \ \ \ \ u^2sin\ u+2ucos\ u-2sin\ u+C\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int (arcsin\ x)^2dx=x(arcsin\ x)^2+2\sqrt{1-x^2}arcsin\ x-2x+C\\\\ &\ \ (22)\ 设u=sin^2\ x，dv=e^xdx，则du=sin\ 2xdx，v=e^x，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^xsin^2\ xdx=e^xsin^2\ x-\int e^xsin\ 2xdx（1-1）求\int e^xsin\ 2xdx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=sin\ 2x，dv=e^xdx，则du=2cos\ 2xdx，v=e^x，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^xsin\ 2xdx=e^xsin\ 2x-2\int e^xcos\ 2xdx（1-2），求\int e^xcos\ 2xdx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=cos\ 2x，，dv=e^xdx，则du=-2sin\ 2xdx，v=e^x，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^xcos\ 2xdx=e^xcos\ 2x+2\int e^xsin\ 2xdx，代入（1-2）式，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^xsin\ 2xdx=\frac{1}{5}e^xsin\ 2x-\frac{2}{5}e^xcos\ 2x+C，代入（1-1）式，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int e^xsin^2\ xdx=e^xsin^2\ x-\int e^xsin\ 2xdx=e^xsin^2\ x-\frac{1}{5}e^xsin\ 2x+\frac{2}{5}e^xcos\ 2x+C=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}e^x-\frac{1}{5}e^xsin\ 2x-\frac{1}{10}e^xcos\ 2x+C\\\\ &\ \ (23)\ 设u=ln\ x，则x=e^u，dx=e^udu，得\int xln^2\ xdx=\int u^2e^{2u}du，\\\\ &\ \ \ \ \ \ \ \ \ \ 设t=u^2，dv=e^{2u}du，则dt=2udu，v=\frac{1}{2}e^{2u}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int u^2e^{2u}du=\frac{1}{2}u^2e^{2u}-\int ue^{2u}du，求\int ue^{2u}du，\\\\ &\ \ \ \ \ \ \ \ \ \ 设t=u，dv=e^{2u}du，则dt=du，v=\frac{1}{2}e^{2u}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int ue^{2u}du=\frac{1}{2}ue^{2u}-\frac{1}{2}\int e^{2u}du=\frac{1}{2}ue^{2u}-\frac{1}{4}e^{2u}+C，代入原式，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\int u^2e^{2u}du=\frac{1}{2}u^2e^{2u}-\int ue^{2u}du=\frac{1}{2}u^2e^{2u}-\frac{1}{2}ue^{2u}+\frac{1}{4}e^{2u}+C，\\\\ &\ \ \ \ \ \ \ \ \ \ \int xln^2\ xdx=\frac{1}{2}x^2ln^2\ x-\frac{1}{2}x^2ln\ x+\frac{1}{4}x^2+C=\frac{1}{2}x^2(ln^2\ x-ln\ x+\frac{1}{2})+C\\\\ &\ \ (24)\ 设u=\sqrt{3x+9}，则x=\frac{1}{3}u^2-3，dx=\frac{2}{3}udu，得\int e^{\sqrt{3x+9}}dx=\frac{2}{3}\int ue^udu，\\\\ &\ \ \ \ \ \ \ \ \ \ 设t=u，dv=e^udu，则dt=du，v=e^u，\\\\ &\ \ \ \ \ \ \ \ \ \ 得\frac{2}{3}\int ue^udu=\frac{2}{3}(ue^u-\int e^udu)=\frac{2}{3}e^u(u-1)+C=\frac{2}{3}e^{\sqrt{3x+9}}(\sqrt{3x+9}-1)+C & \end{aligned} (1) 设u=x，dv=sin xdx，则du=dx，v=−cos x，得∫xsin xdx=−xcos x+∫cos xdx=−xcos x+sin x+C (2) 设u=ln x，dv=dx，则du=x1dx，v=x，得∫ln xdx=xln x−∫dx=xln x−x+C (3) 设u=arcsin x，dv=dx，则du=1−x21dx，v=x，得∫arcsin xdx=xarcsin x−∫1−x2xdx=xarcsin x+1−x2+C (4) 设u=x，dv=e−xdx，则du=dx，v=−e−x，得∫xe−xdx=−xe−x+∫e−xdx=−e−x(x+1)+C (5) 设u=ln x，dv=x2dx，则du=x1dx，v=31x3，得∫x2ln xdx=31x3ln x−31∫x2dx=31x3ln x−91x3+C (6) 设u=cos x，dv=e−xdx，则du=−sin xdx，v=−e−x，得∫e−xcos xdx=−e−xcos x−∫e−xsin xdx，求∫e−xsin xdx，设u=sin x，dv=e−xdx，则du=cos xdx，v=−e−x，得∫e−xsin xdx=−e−xsin x+∫e−xcos xdx，代入原式，得∫e−xcos xdx=−e−xcos x−∫e−xsin xdx=−e−xcos x+e−xsin x−∫e−xcos xdx，则∫e−xcos xdx=21e−x(sin x−cos x)+C (7) 设u=sin 2x，dv=e−2xdx，则du=21cos 2xdx，v=−21e−2x，得∫e−2xsin 2xdx=−21e−2xsin 2x+41∫e−2xcos 2xdx，求∫e−2xcos 2xdx，设u=cos 2x，dv=e−2xdx，则du=−21sin 2x，v=−21e−2x，得∫e−2xcos 2xdx=−21e−2xcos 2x−41∫e−2xsin 2xdx，代入原式，得∫e−2xsin 2xdx=−21e−2xsin 2x+41∫e−2xcos 2xdx=−21e−2xsin 2x−81e−2xcos 2x−161∫e−2xsin 2xdx，则∫e−2xsin 2xdx=−178e−2xsin 2x−172e−2xcos 2xdx+C (8) 设u=x，dv=cos 2xdx，则du=dx，v=2sin 2x，得∫xcos 2xdx=2xsin 2x−2∫sin 2xdx=2xsin 2x+4cos 2x+C (9) 设u=arctan x，dv=x2dx，则du=1+x21dx，v=31x3，得∫x2arctan xdx=31x3arctan x−31∫1+x2x3dx，求∫1+x2x3dx，令u=1+x2，则x=u−1，dx=2u−11du，得∫1+x2x3dx=21∫uu−1du=21∫du−21∫u1du=21u−21ln ∣u∣+C=21(1+x2)−21ln(1+x2)+C，代入原式，得∫x2arctan xdx=31x3arctan x−31∫1+x2x3dx=31x3arctan x−61(1+x2)+61ln(1+x2)+C =31x3arctan x+61ln(1+x2)−61x2+C (10) ∫xtan2 xdx=∫x(sec2 x−1)dx=∫xsec2 xdx−∫xdx，令u=x，dv=sec2 xdx，则du=dx，v=tan x，得∫xsec2 xdx−∫xdx=xtan x−∫tan xdx−21x2，求∫tan xdx， ∫tan xdx=−∫cos x1d(cos x)=−ln ∣cos x∣+C，代入原式，得 xtan x+ln ∣cos x∣−21x2+C (11) 设u=x2，dv=cos xdx，则du=2xdx，v=sin x，得∫x2cos xdx=x2sin x−2∫xsin xdx，求∫xsin xdx，设u=x，dv=sin xdx，则du=dx，v=−cos x，得 ∫xsin xdx=−xcos x+∫cos xdx=−xcos x+sin x+C，代入原式，得∫x2cos xdx=x2sin x−2∫xsin xdx=x2sin x+2xcos x−2sin x+C=(x2−2)sin x+2xcos x+C (12) 设u=t，dv=e−2tdt，则du=dt，v=−21e−2t，得∫te−2tdt=−21te−2t+21∫e−2tdt=−21te−2t−41e−2t+C=−21e−2t(t+21)+C (13) 设u=ln2 x，dv=dx，则du=x2ln xdx，v=x，得∫ln2 xdx=xln2 x−2∫ln xdx，求∫ln xdx，设u=ln x，x=eu，dx=eudu，得∫ln xdx=∫ueudu=xln x−x+C，代入原式，得∫ln2 xdx=xln2 x−2∫ln xdx=xln2 x−2xln x+2x+C (14) 设u=xsin x，dv=cos xdx，则du=(sin x+xcos x)dx，v=sin x，得∫xsin xcos xdx=xsin2 x−∫(sin2 x+xsin xcos x)dx=xsin2 x−∫sin2 xdx−∫xsin xcos xdx，求∫sin2 xdx=−41sin 2x+21x+C，代入原式，得∫xsin xcos xdx=xsin2 x+41sin 2x−21x−∫xsin xcos xdx，则∫xsin xcos xdx=21xsin2 x+81sin 2x−41x+C=−41xcos 2x+81sin 2x+C (15) 设u=cos2 2x，dv=x2dx，则du=−21sin xdx，v=31x3，得∫x2cos2 2xdx=31x3cos2 2x+61∫x3sin xdx（1−1）求∫x3sin xdx，设u=x3，dv=sin xdx，则du=3x2dx，v=−cos x，得∫x3sin xdx=−x3cos x+3∫x2cos xdx（1−2）求∫x2cos xdx，设u=x2，dv=cos xdx，则du=2xdx，v=sin x，得∫x2cos xdx=x2sin x−2∫xsin xdx（1−3）求∫xsin xdx，设u=x，dv=sin xdx，则du=dx，v=−cos x，得∫xsin xdx=−xcos x+∫cos xdx=−xcos x+sin x+C，代入（1−3）式，得∫x2cos xdx=x2sin x−2∫xsin xdx=x2sin x+2xcos x−2sin x+C，代入（1−2）式，得∫x3sin xdx=−x3cos x+3∫x2cos xdx=−x3cos x+3x2sin x+6xcos x−6sin x+C，代入（1−1）式，得∫x2cos2 2xdx=31x3cos2 2x+61∫x3sin xdx=31x3cos2 2x−61x3cos x+21x2sin x+xcos x−sin x+C= 61x3+21x2sin x+xcos x−sin x+C (16) 设u=ln(x−1)，dv=xdx，则du=x−11dx，v=21x2，得∫xln(x−1)dx=21x2ln(x−1)−21∫x−1x2dx，求∫x−1x2dx，设u=x−1，则dx=du，得∫x−1x2dx=∫u(u+1)2du=∫(u+2+u1)du=21u2+2u+ln ∣u∣+C= 21(x+1)2+ln(x−1)+C，代入原式， ∫xln(x−1)dx=21x2ln(x−1)−21∫x−1x2dx=21x2ln(x−1)−41(x+1)2−21ln(x−1)+C= 21(x2−1)ln(x−1)−41x2−21x+C (17) 设u=x2−1，dv=sin 2xdx，则du=2xdx，v=−21cos 2x，得∫(x2−1)sin 2xdx=−21x2cos 2x+21cos 2x+∫xcos 2xdx，求∫xcos 2xdx 设u=x，dv=cos 2xdx，则du=dx，v=21sin 2x，得∫xcos 2xdx=21xsin 2x−21∫sin 2xdx=21xsin 2x+41cos 2x+C，代入原式，得∫(x2−1)sin 2xdx=−21x2cos 2x+21cos 2x+∫xcos 2xdx=−21x2cos 2x+43cos 2x+21xsin 2x+C (18) 设u=ln3 x，dv=x21dx，则du=x3ln2 xdx，v=−x1，得∫x2ln3 xdx=−xln3 x+3∫x2ln2 xdx（1−1），求∫x2ln2 xdx 设u=ln2 x，dv=x21dx，则du=x2ln x，v=−x1，得∫x2ln2 xdx=−xln2 x+2∫x2ln xdx（1−2），求∫x2ln xdx 设u=ln x，dv=x21dx，则du=x1dx，v=−x1，得∫x2ln xdx=−xln x+∫x21dx=−xln x−x1+C，代入（1−2）式，得∫x2ln2 xdx=−xln2 x+2∫x2ln xdx=−xln2 x−x2ln x−x2+C，代入（1−1）式，得∫x2ln3 xdx=−xln3 x+3∫x2ln2 xdx=−xln3 x−x3ln2 x−x6ln x−x6+C (19) 令u=3x，x=u3，则dx=3u2du，得∫e3xdx=3∫u2eudu，设t=u2，dv=eudu，则dt=2udu，v=eu，得3∫u2eudu=3u2eu−6∫ueudu，设t=u，dv=eudu，则dt=du，v=eu，得3∫u2eudu=3u2eu−6∫ueudu=3u2eu−6ueu+6eu+C=3e3x(x32−23x+2)+C (20) 设u=ln x，x=eu，则dx=eudu，得∫cos ln xdx=∫eucos udu，设t=cos u，dv=eudu，则dt=−sin udu，v=eu，得∫eucos udu=eucos u+∫eusin udu=eucos u+eusin u−∫eucos udu， ∫eucos udu=21eu(sin u+cos u)+C=21x(sin ln x+cos ln x)+C (21) 设u=arcsin x，x=sin u，dx=cos udu，得∫(arcsin x)2dx=∫u2cos udu，设t=u2，dv=cos udu，则dt=2udu，v=sin u，得∫u2cos udu=u2sin u−2∫usin udu，设t=u，dv=sin udu，则dt=du，v=−cos u，得∫u2cos udu=u2sin u−2∫usin udu=u2sin u+2ucos u−2∫cos udu)= u2sin u+2ucos u−2sin u+C 得∫(arcsin x)2dx=x(arcsin x)2+21−x2arcsin x−2x+C (22) 设u=sin2 x，dv=exdx，则du=sin 2xdx，v=ex，得∫exsin2 xdx=exsin2 x−∫exsin 2xdx（1−1）求∫exsin 2xdx，设u=sin 2x，dv=exdx，则du=2cos 2xdx，v=ex，得∫exsin 2xdx=exsin 2x−2∫excos 2xdx（1−2），求∫excos 2xdx，设u=cos 2x，，dv=exdx，则du=−2sin 2xdx，v=ex，得∫excos 2xdx=excos 2x+2∫exsin 2xdx，代入（1−2）式，得∫exsin 2xdx=51exsin 2x−52excos 2x+C，代入（1−1）式，得∫exsin2 xdx=exsin2 x−∫exsin 2xdx=exsin2 x−51exsin 2x+52excos 2x+C= 21ex−51exsin 2x−101excos 2x+C (23) 设u=ln x，则x=eu，dx=eudu，得∫xln2 xdx=∫u2e2udu，设t=u2，dv=e2udu，则dt=2udu，v=21e2u，得∫u2e2udu=21u2e2u−∫ue2udu，求∫ue2udu，设t=u，dv=e2udu，则dt=du，v=21e2u，得∫ue2udu=21ue2u−21∫e2udu=21ue2u−41e2u+C，代入原式，得∫u2e2udu=21u2e2u−∫ue2udu=21u2e2u−21ue2u+41e2u+C， ∫xln2 xdx=21x2ln2 x−21x2ln x+41x2+C=21x2(ln2 x−ln x+21)+C (24) 设u=3x+9，则x=31u2−3，dx=32udu，得∫e3x+9dx=32∫ueudu，设t=u，dv=eudu，则dt=du，v=eu，得32∫ueudu=32(ueu−∫eudu)=32eu(u−1)+C=32e3x+9(3x+9−1)+C