練習若F(xf(x,則F(xf(x的原函數(shù)，f(x的原函數(shù)全體稱f的不定積分函數(shù)只是任一個滿足導數(shù)是f(x)的函數(shù)。（1）ex3（2）cosxc（3）1（4）2（5）a（6）1（7） 3.（1）(10xc) ∫10dx10x(2cosxc)sind(x5c)5x

∫2sinxdx2cosx∫5x4dxx5解：由題 f(x) c，又由f(1)1，知c1，因f(x) 1練習6.2

f(x)(lnx)1，所x

f(x)x1.（1）1lnx1 3（2）arcsinx3cosx x32 e

xe1exeex（4）=∫(4x26x9x=ln

4x

6x19x11

（5）=(x2x4x8)dx x8dx x8 2x∫ dxx2xln1xxx2x2∫1x2dx dx∫12xx

1 dx1lnx31ln(x1)

2x

dx

1

x1∫x22x ∫ f(axb)dx1F(axb)∫xxxx

∫1

dx

ln(x2)

4 31333 dx[(1x)(1x)ln1(1x313331

1x∫1x1dx1x∫1(11(1x2

4ln(11x)設tsinx，原式

dttant∫cos2∫11

∫

33原式

3tant，sectdt

∫3tan233x

C解：原式3

dx1lnx (1)2(1)23x29a2a2設xasint，a2sintdta2costa2a214x14x1(x2)2

14xdxlnx14x練習1.∫xlnxdx1∫lnxdx21x2lnx1∫x2dln 1x2lnx1∫xdx1x2lnxx2 ∫x2sinxdx∫x2dcosxx2cosx2∫xcosx2cosx2∫xdsinxx2cosx2xsinx2∫sinx2cosx2xsinx2cosx

xcos

xdcsc2x

xcsc2x

∫sin3 2 2xcsc2x1cotx ∫(1x2)2dx2∫(1x2)2d(1x)2∫xd11 dx 1 2∫1 2∫sec3xdx∫secxdtanxsecxtanx∫tanxdsecsecxtanx∫secxtan2xdxsecxtanx∫sec3xdx∫secsecxtanxlnsecxsecxtanxlnsecxtan∫2

x

xatan

a2tan2tasect

a2x∫a2(sec3tsect)dta2secttantln(seta2x∫2

a

a2xa

xaa2xa2

xa2xlnxxa2xlnx a2x arcsinxdxxarcsinx 1

xarcsinx

1x2∫excosxdxexcosx∫exdcosexcosx∫exsinxdxexcosx∫sinexcosxexsinx∫excos因 ex(cosxsin

∫ecosxdx

t ∫costdt22∫tcos2∫tdsint2tsint2∫sin2tsint2costxx2x xx

exuarctan dx 1arctanu 1arctanu1arctanu∫ d ∫

u11arctanu 1 du1arctanulnu1ln(1u2) ∫u 1 1arctanexlnex

1ln(1e2x)2解：由已知因

f(x)dxsinx∫x∫

f(x)(sinxx∫xfxdx∫xdfxxfx∫f習題1.單選x習題1.單選x 解：已f'(x)因f∫xex∫(x∫udeuueueuc(x)exexc由此曲線過(0,0)知f(x)0.即c1 f(x)xexex1解：由于邊際收入函數(shù) f(x)20知收益函數(shù) R(x)∫(202x)dx20xx2又R(0)0，知c于是R(x)20x需求函數(shù)為Q(x)1005成本為c(x)15Q120.052利潤關于p的函數(shù)為L(x)(1005P)P15(1005P)0.05(1004P)22令L'(P0p

7p120/7時利潤最解：由已知 f

f'(x) f

f(x)ln|x|f(10,c f(x)ln d(3x 33x33x

(3x2)33 (3x2)∫xsinx2dx1sinx2dx21cosx2∫

sincosdxsincos

cosdcoscoscosdcoscos∫excosexdx∫cosexdexsinex dx1∫d(52x2) x dx d(x24x5)1ln|x24x5|∫x24x 2

x24 ∫sin23xdx∫1cos6xdxxsin6x dx∫ dlnx ∫dlnlnxln[ln(lnx)]cx.lnx.ln(lnx) lnxlnlnx lnlnx1x1x22x

dx

ln(u

22d(x(x1)22ln(1xd(x(x1)22∫ecosxsinxdx∫ecosxdcosxecosx∫sin5xdx∫(1cos2x)2dcosx∫(12cos2xcos4x)dcoscosx2cos3x_cos5 dx dln arcsinlnxx3ln2 3ln2 1x ∫d(1arcsinx)1x(1arcsin

sin2xcos2xdx1sin22xdx1(1cos ∫ ∫1x1sin4x (4)2316 1(4)23161x1x∫

dxxx1x1x∫

d(1x2

arcsinx arcsinx∫ ∫ ∫ ∫e

e2x1

arctanex∫tan4xdx∫tan2x(sec2x3∫tan2xdtanx∫tan2xdxtanxtanx

dxcsc4xdxcsc2xdcotx(1cot2x)dcot∫sin4 cot3cotx 3∫sin3xcos5xdx∫(1cos2)cos5xdcosx1cos8x1cos6x

∫ ∫

dx

1dln(t21) 2 ∫∫(11)dtlnt1ct t t

∫t21e1ex1ex∫

x 6t5dt∫6t dtx3 1t 1t36t76t52t36t3ln|t1| t6x6x6x7/66x5/62x3/66x1/66x6x 3secttantx2 ∫9sec2tx21∫costdt1sintc x29 1

x2tan

sec2 3dx∫8sec3t4∫cos(4x2441sint444

dx 1111∫csctdtln|csctcott|

1

tantsec

ln|1 |cln| 11u23∫(2x1)e3xxdx111u23∫1(2x2x)e3x1e3x(4x∫ ∫1(2x2x)e3x1(4x∫ ∫1(2x2x)e3x4x1e3x1e3xd(4x∫ 1(2x2x)e3x4x1e3x4e3x ∫extlnxdx∫exxdxxexex∫excosxdxexsinx∫sinexsinx∫exsinexsinxexcosx∫excos因∫excosxdxsinxcosxex∫2t (ln t 2 2 t

edttedtt e

t2et2∫tetdtt2et2tet2∫t2et2tet2et1((lnx)22lnx2)c ∫lnxdxxlnx∫xdlnxxlnx∫ xlnx22xlnx2x∫e3x

t3∫et3t2dt3∫t2det3t2et3∫etdt3t2et6∫tdet3t2et6tet33x2e3x63xe3x6e3xx dx x21dx2 t1∫x24x ∫x22 ∫t2∫ dt∫t2

dt1lnt291arctantct29 1lnx24x131arctanx2 (35)

2x2x45x2

dx

x2 dx2x 223 223 2

x 2xx2xx22

dx∫ x∫

x21x212

ln1xdx

1

1

1 2

1x

1

1

1x

1

lncos∫cos2x lncosxdtanxtanxlncosx∫

tanxsin tanxlncosxtanxx

x2t 1tu u21du2∫1x23dx2∫1x2 2∫1t 2 u2 11 duu c11

∫1x42dx4∫1x42 1dt dt1ln1t114∫1 4∫1 411ln1t4 41t d2x 1arctan2x1∫2x12 2∫2x12

∫ dx∫ dx

1xex1xex1x2

dxe 11x1x x

x1 t2t242x2tanu

sec

sec3tanu du3tanutanu

3sec2 du∫cscu du3sinucos 3 cotu c

c1 2x 1

e2 ext 41tu∫u41 1dx du41u4141

4 74x3 e1 e

3x24x7dx

x24x7dx

74 x t9x t9t29tt29t29∫ dt t 2

t 93x24x9x212x13x 3x24x9x212x x x1 14x114x1t4arcsin2tcarcsin2x1

3x12dx3x dx∫ x3x dx∫ x16

dx2

x12 3lnx22x171arctanx1 16x28xt2 1 d4x16x28xt241lnt4

t2c