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1、高等数学积分表推导过程 (一)含有 ax+b 的积分 1.Cbaxabaxdbaxadx+=+=+ln1)(11 bax2.()()Cuabaxbaxdadxu uu+=+=+) 1()()(bax1bax 3.Cbaxbbaxabaxbaxd abdaxbaxbax adxbaxbbax adxbaxax adxbaxx+=+=+=+=+)ln(1)(1112224. +=+=+=+baxbaxdbbaxdbbaxdbaxadxbaxbabxbax adxbaxxa adxbaxx)()(2)()(12)(112 32222222Cbaxbbaxbbaxa+ +=ln)(2)(21122 3
2、5.()Cxbax bxCabbaxbCaxbaxbdxaxbaxba baxxdx+=+=+=+=+ln1lnln1ln1lnln111 )(116.()()Cxbabaxba bxxdx ba baxdx badxxbdxbxa baxba xbbaxxdx+=+= +=+lnln11111222222222Cxbax ba bx+=ln127.()()()()Cbaxabbaxabaxdx ab baxdx adxbaxab baxabaxxdx+=+= +=+1ln11122222Cbaxbbaxa+ +=ln128.()()()()()+ +=+=+ =+Cbaxbbaxbaxaba
3、xdx ab baxxdx ab axdxbaxab abxbaxadxbaxx2322222222 222 ln212219.()()()()Cxbax bbaxbCbxbaxbbaxbxdx bbaxdx ba baxbadx baxxdx+=+=+=+ln11lnln1112222222（二）含有bax+的积分 10.()()Cbaxabaxdbaxadxbax+=+=+3 32111.()()()() ()+=+=+=+32325223152 32 521baxbaxaCbaxabbaxadxbaxabdxbaxbaxadxbaxxC+ 12.()()()+=+=+baxaabbaxa
4、dxbaxabdxbaxxabdxbaxbaxadxbaxx231522 722127 322 2 22()()()()CbaxbabxxaaCbaxabbax+=+3222 33 32 3812151052 32 高等数学积分表推导过程 13.()()=+=+=+=+Cbaxabbaxabaxbaxd abdxbaxabaxdx abdxbaxbax abaxxdx23 222 3211()Cbaxbaxa+=232214.()()()()+=+=+baxdbaxabbaxdbaxabaxdx ab baxxdx abdxbaxbax adxbaxx3233222222121()Cbaxbb
5、xxaabaxdx ab+=+222 322 84315215.+baxxdx当 b0 时，有 Cbbaxbbax bbaxdbbaxbbaxbaxba baxxdx+=+ +=+ln111 2当 b+ +0arctan10ln1bCbbax bbC bbaxbbax b16.+=+=+=+badxxbbaxbbaxabxbaxxdx badxbxbaxbaxbaxxdx badxbaxbxbax baxxdx 22 22222222222+=+=+baxxdx ba bxbax baxxdx badxvvuvu baxxdx 22217.+=+=+=+ baxxdxbbaxbaxxdxbdx
6、baxadxbaxxb baxadxxbax2)( 18.+=+=+=+ baxxdx babbxbaxbbaxxdxabaxxdxbdxbaxxb baxxadxxbax 21)(222高等数学积分表推导过程 +=+baxxdxa xbax baxxdxa2（三）含有 x2a2的积分 19.+22axdx设 x=atant(221 时有 ()+naxdx22()()()()()() + + += + +=dx axaaxn axxdx axxn axxnnnnn222122122222122112) 1(2即 ()()()nnnnIaIn axxI2 1122112+ +=于是()()()
7、+ +=1122232121 nnnIn axx naI 由此作递推公式并由 Cax aI+=arctan1 1即可得nI ()()()()()+ += +12221222221232121nnnaxdx annaxanaxdx21.Caxax adxaxaxadxaxaxaxdx=+= +=+=ln2111 211122(四)含有 ax2+b(a0)的积分 22.()+= += +=+0arctan11111222bCxba abxbaxbadba bxbadx bbaxdx()()()0ln2111 212+acbxax的积分 29.dxaacb baxacbxaxdx122244 2 +
8、=+当acb42时有 dxacbabxabacacbxaxdx+ =+424 14422 222令 acbabxa t 4222+ = 则dx acbadt 422= 则原式C acbbaxacbbaxacbtdt aacb acba+ += =4242ln 41 124 44222222综上所述() + +aax的积分 31. +22axdx由于tt22sectan1=+，不妨设+tt， 因此()CaxxCaax axaxdx+=+ += +22 12222lnln 32. ()+322axdx设aax的积分 45. 22axdx当ax 时，设， 由上段结果有()()22 1222222ln
9、lnaxxCauu auduaxdx+=+= = ()CaxxCaaxxC xaxC+=+=+ =+22 12221221lnln1ln 综上所述，Caxx axdx+= 2222ln 高等数学积分表推导过程 46. ()322axdx，设x时有 CxaCatdtttattaaxxdx+=+= arccostansectansec22当0x时，有 高等数学积分表推导过程 ()CxaaxCttadttatdtatdttatatadxxax+=+=arccostan1sectantansecsectan222222当0aax的积分 59.Caxaxdx axadx+= arcsin1122260.
10、 ()322xadx,令+acbxax的积分 73. += +2222421ab ac abxdx acbxaxdx，令tabx=+2，则dtdx = 当042 acb时，则令()0442 22 =uuaacb，则=+= +2222221421utdt aab ac abxdx acbxaxdx再令rutsec=,rdrrudtsectan=,ruuttan22=,于是 += 122tansecln1sec1 tansectan11Crrardradrrurru autdt aacbbaxaacbabxutr 42442sec222+= + = ; acbcbxaxauut rrr 412co
11、scos1tan22222+= 高等数学积分表推导过程 CcbxaxabaxaC acbcbxaxabax acbxaxdx+=+ += +2 122222ln1422ln1当042 acb时，令abxt2+=，uaacb= 242=+dtutadxcbxax222，再令rutsec=，rdrrudttansec= ()Crruarrrruardrruadtuta+=tanseclntanseclntansec21sectan222222 + + =+= acbcbxaxa acbbax aacbaCrruarrua 412 42 24 21tansecln21tansec2122222122
12、Ccbxaxabax abaccbxaxabaxC acbcbxaxabax aacba+=+ +232 2 12222 22ln 84 42422ln44 2当042时，令2244 abacu= 122222222222ln22121Cuttaab aututadt abdt utt adt utabtacbxaxxdx+= = = +Ccbxaxabaxaab acbxaxCacxabxabxaab acxabxa+=+=2212222ln22ln21当0=uab；2batr+=则 ()Cabbaxrudrbatbadt+= =+2arcsin24222282.()()dxxbax；令ta
13、x=；则tax+=；dtdx =；于是()()()+=dtbatbadxxbax2224令02=uab；2batr+=；则()()42arcsin222 2222baxCurururdrrudxxbax=+=()()()Cabbaxabdxxbax+2arcsin82(十一)含有三角函数的积分 83.+=Cxxdxcossin 84.+=Cxxdxsincos 85.+=Cxdxxxxdxcoslncossintan 86.+=Cxdxxxxdxsinlnsincoscot 87.Cx xxdxxxdxxxdxdxxdx+= + = + = + =42tanln42tan42tan42cos42tan422cos2sin22 cossec 2 xxxx xxxx xcotcscsincos1 sin2sin22cos2sin2tan2= ; =+=Cxxxdx2cot2csclnsecCxx+ tansecln 88.CxxCx xxdxxdx xxdx xdxxdx+=+=cotcscln2tanln2tan2tan2cos2tan2cos2
