积分表127个公式的推导(改编打印版)

1、高等数学积 分 表公式推导目录（一）含有（二）含有（三）含有axb 的积分 （ 19）···········································

2、;············ 1axb 的积分 （1018）···································&

3、#183;··············· 5x 2a 2 的积分 （1921）·······························

4、····················· 9（四）含有ax2b (a0) 的积分 （ 2228）························&#

5、183;··················· 11（五）含有（六）含有（七）含有（八）含有ax 2bxc (a0) 的积分 （ 2930）······················

6、83;·················14x 2a 2(a0) 的积分 （ 3144）····························

7、83;············15x 2a 2(a0) 的积分 （4558）·································

8、3;······· 24a 2x 2(a0) 的积分 （5972）······································

9、3;·· 37（九）含有（十）含有a2bxc (a0) 的积分 （ 7378）···································· 48xa或 ( xa )( b x ) 的积分 （7982）

10、3;··························51xb（十一）含有三角函数的积分（ 83112）···················

11、83;······················· 55（十二）含有反三角函数的积分（其中a 0) （ 113121）·····················

12、;··68（十三）含有指数函数的积分（ 122131）·········································· 73（十四）含有对数

13、函数的积分（ 132136）·········································· 78（十五）含有双曲函数的积分（ 137141）·

14、83;········································ 80（十六）定积分（ 142147）······&#

15、183;·················································&#

16、183;··· 81附录：常数和基本初等函数导数公式········································· 85说明 ·&

17、#183;·················································&

18、#183;································· 86（一）含有axb 的积分 （ 19）1.dx1lnaxbCaxb a1b证明：被积函数f ( x )的定义域为 x | xaxba令 axbt (t0) ，则 dtadx ,dx1 dtd

19、x11aax badtt1lntCa将 taxb 代入上式得：dx1 lnax b Cax ba1(ax b 1C(1)2. ( ax b) dxa ( 1)证明：令 axbt ,则dtadx ,dx1 dt1a(axb) dxt dta1t 1Ca ( 1)将tax代入上式得：( axb)1 1Cbdx(ax b)a ( 1)3.xdx12ax bb lnaxbCaxba证明：被积函数f ( x )x的定义域为 x | xbaxba令 ax b t ( t0) , 则 x1 t b , dx1 dtaax1tb11bdxa1axt· dta 2tdtba1dt1b dta 2a 2

20、ttblntCa 2a 21tb lntCa 2将 tax b 代入上式得：xdx1ax b b ln ax bCaxba 24.x 2dx11(axb) 22b (axb)b 2ln axbCax ba 32证明：x2dx1(axb)22abx b2 ) dxaxba2axbb21(axb)dx12abx dx1dxa2a2ax ba 2axb1(axb)dx1(axb)2C1a22a312abx2baxbba 2axdxa3axbd ( ax)b2bdx2b21d (axb)a3a3axb2b3 x22b3lnaxbC2aa1b2dxb 21d(axb)b 2ln axbC3a 2axa3

21、axba3b由以上各式整理得：x 2dx11(axb) 22b (axb) b 2 ln ax bCax ba325.dx1ln axbCx ( ax b)bx1证明：被积函数f ( x )x ( ax设1ABx (ax b)xax bAaB0有Ab1的定义域为 x | xbb)a, 则 1 A( ax b) Bx(A a B) x A b1AbaBb于是dx 1badx11 dxa1dxx (ax b)bx( axb )bxbaxb1111b )bdxbd ( axxax b1ln x1lnax bCbb1lnxC提示：log a b1log a bbaxb1 lnaxbCbx6.dxb)1

22、alnaxbCx2(axbxb 2x证明：被积函数f ( x)1的定义域为 x | xb2(axb)xa设1b)ABCb, 则 1 A x (ax b) B( ax b) Cx 2x2( axxx 2ax即 x 2 (A aC)x ( AbaB)B b1AaAaC0b21有AbaB0BbBb1Ca2b2dxa1dx11a 21dx于是x2( ax b)b2xbx2 dxb2ax ba1 dx11 dxa1d (ax b)b2xbx 2b 2axbalnx1aln axbCb2bxb21alnaxxbCbxb27.xdx1ln ax bbC(ax b) 2a 2ax b证明：被积函数f ( x

23、)x(ax设xA( axb) 2axb即 x A a( AbB)xAa1有AbB0的定义域为 x | xbb) 2aB2 , 则 xA( axb) B(ax b)1AabBa于是xdx11b dxb1dx(ax b)2aaxa(axb) 211b1a 2axb d( axb)a 2(ax b) 2d (ax b)1ba 2 ln axba 2 ( ax b)C1lnaxbbCa 2ax bx 21b 28.( ax b)2 dxa 3 axb2 b lnaxbaxbC证明：被积函数 f ( x )x 2的定义域为 x | xb ( ax b) 2a令 ax b t (t0), 则 x1 t b

24、 , dx1 dtaax2(bt ) 2b 2t 22bt(axb) 2a 2 t 2a 2t 2x 22 dxb2t 22btdtb 21dt1dt2b1( axb)a3t2a3t2a3a3dttb 212ba3ta 3 ta 3 ln tC13 (t2blntb 2) Cat将 taxb 代入上式得：x 2dx1axb2 b lnaxb 2C( ax b)2a3baxb9.dx21b)12 ·ln | ax b | Cx(ax b)b(axbx证明：被积函数f ( x )1的定义域为 x | xbx(axb) 2a设：1ABDx(ax b) 2xaxb(axb)2则1A( ax

25、b) 2Bx (axb)DxAa 2 x 2Ab 22 AabxBax 2BbxDxx 2 ( Aa 2Ba )x(2 AabBbD )Ab 2A1Aa2Ba0b 2a有2 AabBbD0Bb 2Ab 21Dab于是dx11a1a12 dxx(axb)b2dxb2ax bdxb)xb (ax11ln |ax11Cb2 ln |x|b2b|·baxb112·ln | axb |Cb(axb)bx（二）含有axb 的积分 （1018）10.axb dx2(axb) 3C3a111111证明： axb dx(axb) 2 d (axb)(axb) 2C1aa122(axb) 3C

26、3a11.xaxb dx2 2(3ax2b)(axb) 3C15a证明：令 axbt(t0) , 则 xt 2b,dx2tdt, xaxt 2baabtaxaxb dxt 2abt2t dt2(t 4bt 2 )dtaa 22dt52bdt322 t52bt3C5a23a25a3a22t 32(3t 25b)C15a将 taxb代入上式得： xaxb dx2 3(axb)5b (ax b)3C15a 22(3ax2b)(axb) 3C15a 222222312.xax b dx105a 3(15ax12abx8b )(axb)C证明：令 axbt(t0) , 则 xt 2ab,dx2t dt,

27、ax 2axb(t 2b) 2tt 5b2 t2bt 3a2a 2x2axb dx2t(t 5b 2t2bt 3 )dta 32t 6 dt2b 2t 2 dt4b t 4 dta 3a 3a 321t 6 12b 21t 1 24b 1t 4 1Ca 3 1 62a31 2a3 1 44b3 t 523t 72b3t 3C7a3a5a2t 33(15t 435b242bt 2 )C105a将tax代入上式得：bx2axb dx2(axb) 315a 2 x215b 230abx35b242b(axb)105a323(15a 2 x 212abx8b 2 )(axb) 3C105a13.xdx22(ax2b)( axb) Caxb3a证明：令 axbt(t0) , 则 xt 2b,dx2t dt,aaxdxt 2b2t dtaxbata2t 2 dt2bdta 2a 22112t 212btCa 2a 22t32bC3a2a2t将tax代入上式得：xdx22( axb)(axb)2b(ax b) Cbaxb3aa22(ax2b)( axb)C3a214.x 2dx23(3a2 x 24abx8b2 )( axb)Caxb15a证明：令 axbt(t0) , 则 xt 2b,dx2t dt ,x 2(t 2aabdxb) 2 1 2t dtaxata2(t 4b