.
dxexdxdexx???arctane?C (23) ?x?x2x2x??e?ee?1e?12x?3d(x2?3x?8)dx??2?ln|x2?3x?8|?C (24) ?2x?3x?8x?3x?8x2?2(x?1)2?2(x?1)?3dx??dx(25) ?33(x?1)(x?1)
12323??(??)dx?ln|x?1|???C232x?1(x?1)x?12(x?1)(x?1)(26)
?dxdxx?a22
解 令x?atant, 则
?asec2tdt???ln|sect?tant|?C1?ln|x?x2?a2|?C
asectx2?a2dx1x1x?d??(x2?a2)32a2?(x2?a2)12a2(x2?a2)12?C
(27)
解法2 令x?atant, 则
dxasec2tdt11x??costdt?sint?C??C 2?(x2?a2)32?a3sec3ta2?222aax?a(28)
?x5x51?x2dx
解 令x?sint, 则
?sin5tcostdx??dt??sin5tdt???(1?cos2t)2dcostcost1?x2123252
2121??cost?cos3t?cos5t?C??(1?x2)?(1?x2)?(1?x2)?C3535(29)
?1?16x3xdx
65解 令x?t, 则x?t, dx?6t
'.
.
t3?t5dt(t2)4?1?1(t2?1)(t6?t4?t2?1)?1dt2?1?3xdx?6?1?t2?6?1?t2dt?6?1?t
7531ttt6t?1?6?(?(t6?t4?t2?1)?)dt??6(???t)?ln||?C27532t?11?tx其中t?x (30)
16?x?1?1x?1?1dx
2解 令x?1?t, 则x?1?t, dx?2tdt,
?x?1?1x?1?1dx??t?124t42tdt??(1?)2tdt??(2t?)dt??(2t?4?)dtt?1t?1t?1t?1
?t2?4t?4ln|t?1|?C1?x?1?4x?1?4ln|x?1?1|?C1?x?4x?1?4ln|x?1?1|?C2．应用分部积分法求下列不定积分： ⑴
?arcsinxdx?xarcsinx??x1?x2dx?xarcsinx?1?x2?C
⑵
1lnxdx?xlnx?x???xdx?xlnx?x?C
222xcosxdx?xdsinx?xsinx??2xsinxdx??⑶
?x2sinx??2xdcosx?x2sinx?2xcosx?2?cosxdx ?x2sinx?2xcosx?2sinx?C⑷ ⑸
lnx?11?1lnx11lnx1dx?lnxd??dx????C ?x32?2x22?x3x22x24x2222(lnx)dx?x(lnx)?2lnxdx?x(lnx)?2xlnx?2x?C ??1121x22dx ⑹ ?xarctanxdx??arctanxdx?xarctanx??22221?x?1211121xarctanx??(1?)dx?xarctanx?(x?arctanx)?C 22221?x211?(x2?1)arctanx?x?C 22'.
.
⑺ [ln(lnx)?11]dx?ln(lnx)dx????lnxdx lnx11?xln(lnx)??x?dx??dx?xln(lnx)?C
xlnxlnx22(arcsinx)dx?x(arcsinx)???⑻
2xarcsinx1?x2dx
?x(arcsinx)2?2?arcsinxd1?x2 ?x(arcsinx)2?21?x2arcsinx?2?1?x211?x2dx
?x(arcsinx)2?21?x2arcsinx?2x?C
⑼
?sec3xdx??secxdtanx?secxtanx??secxtan2xdx
?secxtanx??secx(sec2x?1)dx?secxtanx??sec3xdx??secxdx ?secxtanx??sec3xdx?ln|secx?tanx|
所以
3sec?xdx?1secxtanx?ln|secx?tanx|)?C 2xx?a22⑽
?x2?a2dx?xx2?a2??x?2222dx
?xx?a??(x?a?2222a2x?a22)dx
?xx?a??x?adx??a2x?a22dx
?xx2?a2??x2?a2dx?a2ln(x?x2?a2)
所以
?x2?a2dx?1(xx2?a2?a2ln(x?x2?a2))?C 2类似地可得
?'.
x2?a2dx?1(xx2?a2?a2ln(x?x2?a2))?C 23．求下列不定积分：
.
⑴ [f(x)]f?(x)dx?[f(x)]df(x)?⑵
?a?a1[f(x)]a?1?C a?1f?(x)1dx??1?[f(x)]2?1?[f(x)]2df(x)?arctanf(x)?C
⑶
?f?(x)df(x)dx???ln|f(x)|?C f(x)f(x)⑷ ef(x)f?(x)dx?ef(x)df(x)?ef(x)?C
??4．证明：
⑴ 若In?tannxdx，n?2,3,?，则In??1tann?1x?In?2 n?1证 In?tann?2x(sec2x?1)dx?tann?2xsec2xdx?tann?2xdx
?????tann?2xdtanx?In?2.
因为tann?2xdtanx?tann?1x?(n?2)tann?2xdtanx，
??所以tan从而In??n?2xdtanx?1tann?1x. n?11tann?1x?In?2. n?1⑵ 若I(m,n)?cosmxsinnxdx，则当m?n?0时，
?cosm?1xsinn?1xm?1I(m,n)??I(m?2,n)
m?nm?ncosm?1xsinn?1xn?1???I(m,n?2)，n,m?2,3,?
m?nm?n证 I(m,n)?cosxsinxdx??mn1m?1n?1cosxdsinx ?n?1?1[cosm?1xsinn?1x?(m?1)?cosm?2xsinn?2xdx] n?11?[cosm?1xsinn?1x?(m?1)?cosm?2xsinnx(1?cos2x)dx] n?11?[cosm?1xsinn?1x?(m?1)(I(m?2,n)?I(m,n))] n?1'.
.
cosm?1xsinn?1xm?1?I(m?2,n)， 所以I(m,n)?m?nm?ncosm?1xsinn?1xn?1?I(m,n?2) 同理可得I(m,n)??m?nm?n
P.199 习题
1．求下列不定积分：
x3x3?1?11dx??dx??(x2?x?1?)dx ⑴ ?x?1x?1x?1x3x2???x?ln|x?1|?C 32x?221(x?4)2dx??(?)dx?ln?C ⑵ 解法一：?2x?4x?3|x?3|x?7x?12解法二：
x?212x?713dx?dx?dx 22?x2?7x?12??2x?7x?122x?7x?121d(x2?7x?12)3??2??22x?7x?1217d(x?)
712(x?)2?24?13x?4ln|x2?7x?12|?ln?C 22x?311ABx?C??? 3221?x(1?x)(1?x?x)1?x1?x?x2⑶ 解
去分母得 1?A(1?x?x)?(Bx?C)(1?x)
令x??1，得A?13. 再令x?0，得A?C?1，于是C?23. 比较上式两端二次幂的系数得 A?B?0，从而B??13，因此
dx1dx1x?2???1?x33?1?x3?1?x?x2dx'.
数学分析课后习题答案(华东师范大学版)
