?2x?1?arctan(2x?1) 参见教材P.186 例9或P.193关于Ik的递推公式⑺. 2(2x?1)?1于是，有
x?21152x?15dx????arctan(2x?1)?C ?(2x2?2x?1)242x2?2x?12(2x?1)2?12?5x?35?arctan(2x?1)?C 22(2x?2x?1)2
2．求下列不定积分
dx?5?3cosx
x解 令t?tan，则
2⑴
dx?5?3cosx??dx2dtdt1d(2t)1???arctan2t?C 2222??21?(2t)21?t1?t1?4t5?31?t2?1xarctan(2tan)?C 22⑵
dxdxdxdtanx????2?sin2x?2cos2x?3sin2x?(2?3tan2x)cos2x?(2?3tan2x)
d(3tanx)1132??arctan(tanx)?C ?326(1?tan2x)62⑶
dxcosxdx1cosx?sinx?sinx?cosx??dx ?1?tanx?cosx?sinx2?cosx?sinx1?sinx?cosx1d(sinx?cosx)??(1?)dx?(?dx??) 2cosx?sinx2cosx?sinx1?(x?ln|cosx?sinx|)?C 2cosxdxsinxdx另解：设I1??，I2??，
cosx?sinxcosx?sinx 167
cosx?sinx?cosx?sinxdx?x?C， cosx?sinxd(cosx?sinx)I1?I2??dx???ln|cosx?sinx|?C
cosx?sinxcosx?sinxdx1所以??I1?(x?ln|cosx?sinx|)?C
1?tanx2则I1?I2?⑷
?x21?x?x2dx???1?x?x2dx??(x?1)dx1?x?x2
???1?x?x2dx?1(?2x?1)dx3dx ?2?1?x?x22?1?x?x2其中（利用教材P.185例7的结果）
?1?x?x2dx??51152x?11?(x?)2dx?[arcsin?(x?)1?x?x2] 422425??(?2x?1)dx1?x?x2dx1?x?x2????d(1?x?x2)1?x?x2dx?21?x?x2
2x?1551?(x?)242?arcsin
所以有
?x21?x?x2dx
152x?11132x?1??[arcsin?(x?)1?x?x2]?21?x?x2?arcsin?C
2422255?72x?12x?3arcsin?1?x?x2?C 845dxx?x2⑸
???1d(x?)12?ln|x??x2?x|?C 211(x?)2?24 168
⑹
11?x?x21?xdx
1?t21?x?4tdt解 令 t?，则x?，，代入原式得 dx?2221?t1?x(1?t)1?x2?1?t21?xdx????1?t21?x???4t?4t21?t2?1???t?(1?t2)2dt??(1?t2)2dt?4?(1?t2)2dt ?2?4??2?111112dt?4dt?4dt?[??]dt 2222222???1?t(1?t)1?t(1?t)(1?t)1?t1111?t11dt?[?]dt?ln||???C 222?1?t1?t1?t1?t(1?t)(1?t)1?1?x21?x2?ln||??C
xx
总 练 习 题
求下列不定积分： ⑴
14112145133?x?23x?14xdx??(x?2x4241244?x)dx?x4?x?x?C
5133?⑵
?xarcsinxdx?x2112122arcsinxdx?[xarcsinx?xdx] ??2221?x其中
??sin2t1?cos2t11dx??costdt??dt?(t?sin2t)
cost2221?x21(arcsinx?x1?x2) 2所以xarcsinxdx??121[xarcsinx??x2dx]
221?x11?[x2arcsinx?(arcsinx?x1?x2)]?C 22
169
?1211xarcsinx?arcsinx?x1?x2?C 244⑶
?1?dxx
解 令x?u，则dx?2udu
?1?dxx??2udu1?2?(1?)du?2(u?ln|1?u|)?C 1?u1?u?2(x?ln|1?x|)?C
⑷
?esinxsin2xdx?2?esinxsinxcosxdx?2?esinxsinxdsinx?2?sinxdesinx
?2(esinxsinx??esinxdsinx)?2(esinxsinx?esinx)?C?2esinx(sinx?1)?C
⑸
?e?xxdx(令x?u)?eu2udu?2(euu?eu)?C?2ex(x?1)?C
⑹
dxx2?1??dxx21?1x2???11d()??arcsin?C xx11?2x1解法二：令x?sect，
?x⑺
dxx2?1??secttant1dt?t?C?arccos?C
secttantx1?tanxcosx?sinxd(cosx?sinx)dx?dx??1?tanx?cosx?sinx?cosx?sinx
?ln|cosx?sinx|?C
1?tanx??dx?tan(?x)dx?ln|cos(?x)|?C ?1?tanx?44x2?x(x?2)2?3(x?2)?231⑻ ?dx?dx?ln|x?2|???C 332?x?2(x?2)(x?2)(x?2)dxdx122?secx?(1?tanx)dtanx?tanx?tan3x?C 2?cos4x??3cosx1?cos2x2422)dx ⑽ ?sinxdx??(sinx)dx??(2⑼
170
111?cos4x2(1?2cos2x?cos2x)dx?(1?2cos2x?)dx 4?4?21xsin4x311?(x?sin2x??)?C?x?sin2x?sin4x?C 4288432x?5dx ⑾ ?32x?3x?4?解
x?5x?5dx??x3?3x2?4?(x?1)(x?2)2dx
令
x?5ABC ???22x?1x?2(x?1)(x?2)(x?2)去分母得：x?5?A(x?2)2?B(x?1)(x?2)?C(x?1) 解得：A??所以
22，B?，C??1 33x?521211dx??dx?dx??x3?3x2?4?(x?2)2dx 3?x?13?x?2?2x?21ln||??C 3x?1x?2⑿
?arctan(1?x)dx
解 令1?x?u，dx?2(u?1)du
x)dx??arctanu?2(u?1)du?2?arctanu?udu?2?arctanudu
?arctan(1??[(u2?1)arctanu?u]?2uarctanu?ln(1?u2)?C1
?xarctan(1?x)?x?ln(2?x?2x)?C
x7x7?2x3?2x32x33dx??dx??(x?4)dx ⒀ ?4x?2x4?2x?2?141x?ln(x4?2)?C 42x7x4?x31211dx??4dx??(1?4)dx4?x4?ln(x4?2)?C 另解：?4442x?2x?2x?2
171