=secxtanx?
3sec?xdx?23(secx?1)secxdx?secxtanx?ln|secx?tanx|?sec??xdx
11secxtanx?ln|secx?tanx|?c 223.
(lnx)3?x2dx
(lnx)3113(lnx)233?x2dx???(lnx)dx??x(lnx)??x2dx
(lnx)33(lnx)26lnx(lnx)33(lnx)26lnx6?????2dx??????2dx
xxxxxxx(lnx)33(lnx)26lnx6??????c
xxxx解.
4.
?cos(lnx)dx
?cos(lnx)dx?xcos(lnx)??sin(lnx)dx?x[cos(lnx)?sin(lnx)]??cos(lnx)dx
解.
x[cos(lnx)?sin(lnx)]?c 2xxxcos4xcos42dx?12dx??1xdsin?2x??1xsin?2x?1sin?2xdx
5. ?8?3x8?2828?2sin3x3xsincos2211xx1x1x?2x ??xsin??sin?2d??xsin?2?cot?c 824228242 ?
?cos(lnx)dx?六. 求下列不定积分:
1.
?xln(x?1?x2)dx 22(1?x)xln(x?1?x2)112dx?ln(x?1?x)d(1?x2)22?1?x211111ln(x?1?x2)??dx 22?221?x21?x1?x
解.
? =
ln(x?1?x2)1112???sectdt 令x?tant 22?2(1?x)21?tantsectln(x?1?x2)1cost =?dt 22?2(1?x)21?2sintln(x?1?x2)1d2sint = ?22?2(1?x)221?2sintln(x?1?x2)11?2sint =?ln?c
2(1?x2)421?2sintln(x?1?x2)11?x2?2x =?ln?c 222(1?x)421?x?2x2.
?xarctanx1?x2dx
解.
?xarctanx1?x2dx??arctanxd1?x?1?xarctanx??11?x2221?x2dx
1?x2 =
1?x2arctanx??dx?1?x2arctanx?ln(x?1?x2)?c
3.
arctanex?e2xdx
arctanex11?2x1?2xexx?2xx?e2xdx??2?arctanede??2earctane?2?e1?e2xdx 1?2x1e?x1?2x11xx??earctane??dx??earctane?dx 2xx2x?221?e22e(1?e)解.
1?2x11ex1?2xxx?x??earctane??(x?)dx??(earctane?e?arctanx)?c 2x22e1?e2七. 设
?xln(1?x2)?3x?0f(x)??2 , 求?f(x)dx.
?xx?0?(x?2x?3)e?(xln(1?x2)?3)dx??f(x)dx??
2?x(x?2x?3)edx???解.
?
12?122xln(1?x)?[x?ln(1?x2)]?3x?cx?0? ??22x?02?x??(x?4x?1)e?c?1考虑连续性, 所以
c =－1+ c1, c1 = 1 + c
?12?122xln(1?x)?[x?ln(1?x2)]?3x?cx?0?f(x)dx??2 2x?02?x??(x?4x?1)e?1?c?f'(ex)?asinx?bcosx, (a, b为不同时为零的常数), 求f(x).
八. 设
解. 令t
?ex，x?lnt, f'(t)?asin(lnt)?bcos(lnt), 所以
f(x)??[asin(lnx)?bcos(lnx)]dx
=
x[(a?b)sin(lnx)?(b?a)cos(lnx)]?c 2九. 求下列不定积分: 1.
x3?2?3x(2x?3)dx
22解.
3x?3xx?3xx?3x2?3(2x?3)dx??3d(x?3)?ln3?c
22.
?(3x2?2x?5)(3x?1)dx
32332解.
12(3x?2x?5)(3x?1)dx?(3x2?2x?5)2d(3x2?2x?5) ??21?(3x2?2x?5)2?c 55
3.
?ln(x?1?x2)1?x2dx
12ln(x?x2?1)?c 2解.
?ln(x?1?x2)1?x2dx??ln(x?x2?1)dln(x?x2?1)?
4.
?(1?xxdx2?x?1)ln(1?x?1)xdx222解.
?(1?x?x2?1)ln(1?x2?1)??dln(1?x2?1)ln(1?x2?1)?ln|ln(1?x2?1)|?c
十. 求下列不定积分: 1.
xarctanx?(1?x2)dx
xarctanx1arctanx122?1dx?d(1?x)??arctanxd(1?x) 22?(1?x2)2??2(1?x)2??1arctanx111arctanx11?darctanx???dx
21?x22?1?x221?x22?(1?x2)2解.
令x?tant?1arctanx11arctanx11?cos2t2?costdt????dt
21?x22?21?x2221arctanx111aextanx11???t?sin2t?c???arctanx?sintcost?c
21?x24821?x2441aextanx11x???arctanx??c 2221?x441?xxdx 1?xx?t,1?x则x?tan2t
2.
?arcsin解. 令arcsin
?arcsinxdx??tdtan2t?ttan2t??tan2tdt?ttan2t?tant?t?c 1?xxxx?x?arcsin?c?(1?x)arcsin?x?c 1?x1?x1?x
?xarcsin3.
arcsinx1?x2?x2?1?x2dx解.
arcsinx1?x2?x2?1?x2dx令x?sintt1?sin2t2?sin2t?costcostdt??t(csct?1)dt
1???tcottdt??tdt??tcott??cottdt?t2?c
21??tcott?ln|sint|?t2?c
21?x21??arcsinx?ln|x|?(arcsinx)2?c
x2
4.
arctanx?x2(1?x2)dx
arctanx?x2(1?x2)dx令x?tantt22sectdt?t(csct?1)dt ?tan2tsec2t?解.
11??tcsc2tdt??tdt???tdcott?t2??tcott??cotdt?t2
22
1arctanxx1??tcott?ln|sint|?t2?c???ln||?(arctanx)2?c
2x21?x2
arctanx1x212???ln?(arctanx)?c 2x21?x2十一. 求下列不定积分: 1.
32x4?xdx ?32332x4?xdx令x?2sint8sint2cost2costdt?32sintcostdt ???解.
?32?(1?cos2t)cos2tdtdcost??41??(4?x2)2?(4?x2)2?c
35353232cos3t?cos5t?c 35
2.
??x2?a2x
解.
x2?a2令x?asectxatant1?cos2t?asectasecttantdt?a?cos2tdt
a?atant?at?c?x2?a2?aarccos?c
x2x3.
?ex(1?ex)1?edx
t(1?t)dt1?t1?sinu?dt令t?sinu?1?t2t?1?t2?cosucosudu
解.
?ex(1?ex)1?e2xdx令ex?t
?u?cosu?c?arcsinex?1?e2x?c
xdx (a > 0)
2a?x4.
?x解.
xu424xdx8asin 令u?x令u?2asint2du?2a?x?tdt ?2a?u22(1?cos2t)2dt?2a2?(1?2cos2t?cos22t)dt =8a?41?cos4ta422dt?3at?2asin2t?sin4t?c =2at?2asin2t?2a?24222 =3a =3a2t?4a2sintcost?a2sintcost(1?2sin2t)?c
2t?3a2sintcost?2a2sin3tcost?c arcsinxx2a?xx?3a2?2a22a2a2a2ax3a?x?2a2x(2a?x)?c
=3a2x2a?x?c
2a2a =3a2arcsin十二. 求下列不定积分: 1.
?sinxdx1?cosx
解.
?sinxdx1?cosx??sinxdxsin2x1?cosx???d(1?cosx)sin2x1?cosx??2?d1?cosx
1?cos2x
令1?cosx?u?2?dudu??2?u2(2?u2)
1?(u2?1)2
???(?11112?u?)du??ln||?c 22u22u2?u2?u1?122ln|2?1?cosx2?1?cosx|?c
1?cosx2.
2?sinx?2?cosxdx
高数考研习题及答案
