Transcription of Formulario de integrales - UV
1 Formulariodeintegralesc 2001-2005 SalvadorBlascoLlopisEsteformulariopuedes ercopiadoy distribuidolibrementebajolalicenciaCreat iveCommonsAtribuci ~ eptimarevisi on:Febrero2005 Sextarevisi on:Julio2003 Quintarevisi on:Mayo 2002 Cuartarevisi on:Mayo 2001 Tercerarevisi +b(1)Z(ax+b)ndx=1a(n+ 1)(ax+b)n+1+C; n6= 1(2)Zdxax+b=1alnjax+bj+C(3)Zdxx(ax+b)=1a ln xax+b +C(4)Zdx(1 + x)2= 1 11 + x+C(5)Zxdx(1 +bx)3= 12b 2(1 +bx)2 12b 11 +bx+ +b(6)Zxpa+bxdx=2(3bx 2a)(a+bx)3=215b2+C(7)Zxpa+bxdx=2(bx 2a)pa+bx3b2+C1(8)Zdxxpa+bx=8<:1paln pa+bx papa+bx+pa +C; a >02p aarctanqa+bx a+C; a <0(9)Zpa+bxxdx= 2pa+bx+aZdxxpa+bx+ a2(10)Zdxa2+x2=1aarctanxa+C; a >0(11)Zxdx(x2 a2)3=2=1px2 a2+ x2; x < a(12)Z(a2 x2)3=2dx=x2pa2 x2+a22arcsenxa+C; a >0(13)Zdxa2 x2=12aln a+xa x +C=1aarctanhxa(14)Zdx(a2 x2)3=2=xa2pa2 x2+ a2 Zpx2 a2dx=12 xpa2 x2+a2ln x+pa2 x2 +C=(15)=(12xpa2+x2+a22arcsenhx+C(+)12xpa 2 x2+a22arccoshx+C( )(16)Zxpx2 a2dx=13(x2 a2)3=2+C(17)Zx3px2+a2dx= (15x2 25a2)(a2+x2)3=2+C(18)Zpx2 a2xdx=px2 a2 a arccosajxj+C(19)Zdxpx2+a2=a arcsenhxa+C(20)Zdxpx2 a2= ln x+px2 a2 +C= arccoshxa+C;(a >0)(21)Zdxxpx2 a2=1aarccosajxj+C;(a >0)2(22)Zdxx2px2 a2= px2 a2a2x+C(23)Zxdxx2 a2=px2 a2+C(24)Zpx2 a2x4dx= (a2+x2)3=23a2x3+C(25)Zx2px2 a2dx=x2px2 a2 a22arccoshxa+ x2(26)Zpa2 x2dx=12xpa2 x2 a22arcsenxa+C;(a >0)(27)Zxpa2 x2dx= 13(a2 x2)3=2+C(28)Zx2pa2 x2dx=x8(2x2 a2)pa2 x2+a28arcsenxa+C; a >0(29)Zpa2 x2xdx=pa2 x2 aln a+pa2 x2x +C(30)Zdxx2pa2 x2= pa2 x2a2x+C(31)Zdxpa2 x2= arcsenxa+C; a >0(32)Zdxxpa2 x2= 1aln a+pa2 x2x +C(33)Zxpa2 x2dx= pa2 x2+C(34)Zx2pa2 x2dx= x2pa2 x2 a22arcsenxa+C; a >0(35)Zdxpa2+x2= ln x+pa2+x2 +C= arcsenhxa+C.)
2 A > +bx+c(36)Zdxax2+bx+c=8>>>> <>>>>:1pb2 4acln 2ax+b pb2 4ac2ax+b+pb2 4ac ==2pb2 4acarctanh2ax+bpb2 4ac+C; b2>4ac2p4ac b2arctan2ax+bp4ac b2+C; b2<4ac 22ax+b+C;b2= 4ac3(37)Zxax2+bx+cdx=12aln ax2+bx+c b2aZdxax2+bx+c+C(38)Zx dx(ax2+bx+c)n=bx+ 2c(b2 4ac)(n 1)(ax2+bx+c)n 1++b(2n 3)(b2 4ac)(n 1)Zdx(ax2+bx+c)n 1; n6= 0;1; b2<4ac(39)Zdx(ax2+bx+c)n=2ax+b (b2 4ac)(n 1)(ax2+bx+c)n 1++2a(2n 3) (b2 4ac)(n 1)Zdx(ax2+bx+c)n 1; n6= 0;1; b2< +bx+c(40)Zpax2+bx+cdx=2ax+b4apax2+bx+c+4ac b28aZdxpax2+bx+c(41)Za0+a1x+:::+anxnpax2+bx+ ,p :m etodoalem an(42)Zdxpax2+bx+c=1paln 2ax+b+papax2+bx+c +C=8> <>:1paarcsenh2ax+bp4ac b2+C; <0;a >0;1palnj2ax+bj+C; = 0;a >0;1 p aarcsen2ax+bpb2 4ac+C; >0;a <0;; =b2 4ac(43)Zxpax2+bx+cdx=pax2+bx+ca b2aZdxpax2+bx+c(44)Zdxxpax2+bx+c=8<: 1pcln 2pcpax2+bx+c+bx+2cx +C; c >01p carcsenbx+2cjxjpb2 4ac+C;c < (45)Zdxsenax=1aln tanax2 +C(46)Zsen2axdx=12 ax cosax senaxa+C=x2 sen2ax4a+C4(47)Zsennaxdx= senn 1ax cosaxa n+n 1nZsenn 2axdx;n6= 0; 1;(48)Zsenaxsenbxdx= 12cos(a+b)xa+b 12cos(a b)xa b+C; a26=b2(49)Zxnsenaxdx= 1axncosax+naZxn 1cosaxdx(50)Zsenaxxdx=1X =0(ax)2 1(2 1) (2 1)!
3 (51)Zdx1 senax=1atan ax2 4 + (52)Zcos2axdx=12 ax+ cosax senaxa+C(53)Zdxcosax=1aln tan ax2 4 +C(54)Zcosaxxdx= lnjaxj+1X =1( 1) (ax)2 (2 ) (2 )!+C(55)Zcosnaxdx=cosn 1ax senaxa n+n 1nZcosn 2axdx+C; n6= 0; 1;(56)Zcosaxcosbxdx= 12sen(a b)xa b+12sen(a+b)xa+b+C; a26=b2(57)Zxncosaxdx=1axnsenax naZxn 1senaxdx+C; n6= 1(58)Zdx1 cosax= 1atanax2+ (59)Ztanaxdx= 1aln cosax+C(60)Ztan2xdx= tanx x+C(61)Ztannaxdx=tann 1axa(n 1) Ztann 2axdx;n6= 1;0;(62)Zcotaxdx=1aln senax+C(63)Zcotnaxdx= cotn 1axa(n 1) Zcotn 2axdx+C; n6= 1;0; (64)Zsecaxdx=1ahln cosax2+ senax2 ln cosax2 senax2 i+C(65)Zsec2axdx=1atanax+C(66)Zsecnxdx=1 atanax secn 2axn 1+1an 2n 11secn 2ax dx+C; n6= 1;(67)Zcsc2axdx= 1acotax+C(68)Zcscnaxdx= 1acotax cscn 2n 1+1a n 2n 1 Zcscn 2ax dx+C; n6= 1; (69)Zsecx tanax dx= secx+C(70)Zcscx cotx dx= cscx+C(71)Zcosmx sennx dx=(cosm 1x senn+1xm+n+m 1m+nRcosm 2x sennx dxcosm+1x senn 1xm+n+n 1m+nRcosmx senn 2x etricasinversas(72)Zarcsenxadx=x arcsenxa+pa2 x2+C; a >0;(73)Zarccosxadx=x arccosxa pa2 x2+C; a >0;(74)Zarctanxadx=x arctanxa 1aaln 1 +x2a2 +C; a >0.)
4 (75)Zarccotxadx=1ax arccotxa+a2ln(a2+x2) +C(76)Zx arccosxdx=x arccosx+ arcsenx+ tmicas(77)Zxneaxdx=xneaxa naZxn 1eaxdx(78)Zeaxsenbx dx=eax(asenbx bcosbx)a2+b2+C(79)Zeaxcosbx dx=eax(bsenbx+acosbx)a2+b2+C(80)Zdxa+ben x=xa ln(a+benx)an+C(81)Zlogaxdx=xlogax xlna+C;8a >0;(82)Zxlnxdx=2x2lnx x24+C(83)Zxnlnax dx=xn+1 lnaxn+ 1 1(n+ 1)2 +C(84)Zxn(lnax)mdx=xn+1n+ 1(lnax)m mn+ 1 Zxn(lnax)m 1dx; n;m6= 1;x >0;(85)Zlnaxdx=xlnax x+C; x >0;(86)Zeaxxdx= lnjxj+1Xi=1(ax)ii i!+C(87)Zeaxlnx dx=1aeaxlnjxj 1aZeaxxdx+C(88)Zdxlnx= lnjlnxj+1Xi=1lnixi i!+C; x >0;(89)Zdxxlnx= lnjlnxj+C; x >0;(90)Zlnnxxdx=1x+ 1lnn+1x; n6= 1;x >0; olicas(91)Zsenhaxdx=1acoshax+C(92)Zsenh2 xdx=14senh2x 12x+C(93)Zcoshaxdx=1asenhax+C(94)Zcosh2x dx=14senh2x+12x+C(95)Ztanhaxdx=1alnjcosh axj+C(96)Zcothaxdx=1alnjsenhaxj+C(97)Zse chxdx= arctan(senhx) +C(98)Zsech2xdx= tanhx+C(99)Zcschxdx= ln tanhx2 = 12lncoshx+ 1coshx 1+C(100)Zsenhx tanhx dx= sechx+C(101)Zcschx cothx dx= cschx+ olicasinversas(102)Zarcsenhxadx=xarcsenh xa pa2+x2+C; a >0(103)Zarccoshxadx=(xarccoshxa px2 a2+C;arccoshxa>0;a >0;xarccoshxa+px2 a2+C;arccoshxa<0;a >0;(104)Zarctanhxadx=xarctanhxa+12aln(a2 x2) +C(105)Zarccothxadx=xarccothxa+12aln(x2 a2) +C(106)Zarcsenhxadx=xarcsechxa aarcsenr1 x2a2+C(107)Zarcsechxadx=xarccschxa+aarcc oshr1 +x2a2+ nidas(108)Z10xne qxdx=n!)
5 Qn+1; n > 1;q >0;Z10xme ax2dx= [(m+ 1)=2]2a(m+1)=2; a >0(109)=n!2an+1;Simimpar:m= 2n+ 1=1 3 ::: (2n 1)2n+1r a2n+1;Simpar:m= 2n(110)Z 0x2e ax2dx= 2ae a 2+14r a3erf( pa)(111)Z1txne axdx=n!e atan+1 1 +at+a2t22!+:::+antnn! ; n= 0;1;:::;a >0;(112)Z10xne ax2dx=n 12aZ10xn 2e ax2dx(113)Z10e ax2dx=12r a(114)Z10xe ax2dx=12a(115)Zx0dx1 x= ln11 x(116)Zx0dx(1 x)2=x1 x(117)Zx0dx1 + x=1 ln(1+ x)(118)Zx01 + x1 xdx= (1+ ) ln11 x x(119)Zx01 + x(1 x)2dx=(1 )x1 x ln11 x(120)Zx0(1 + x)2(1 x)2dx= 2 (1 + ) ln(1 x) + 2x+(1 + )2x1 x(121)Zx0dx(1 x)( B x)=1 B 1ln B x B(1 x); B6= 1(122)Zx0dxax2+bx+c= 22ax+b+2b; b2= 4ac(123)9Zx0dxax2+bx+c=1a(p q)ln qp x px q ; b2>4ac;p;qsonlasra ces;(124)Zx0a+bxc+gxdx=bxg+ag bcg2ln(c+gx) etodosdeintegraci onporpartes:Zu dv=u v Zv onporsustituci on:six=g(t) esunfunci onqueadmitederivadacont nuanonulay funci oninversat=h(x) yF(t) esunaprimitiva def(g(t))g0(t) setieneque:Zf(x)dx=F(h(x)) + ondefuncionesracionales:QueremoshallarRF (x)Q(x)dxsiendoF(x) yQ(x) polinomiosdecoe gradodeFesmayorqueel deQsehaceladivisi onparaobtenerRF(x)Q(x)dx=RC(x)dx+RR(x)Q( x)dx.
6 (x) sepuededescomponerdelasiguientemanera:Q( x) =a0(x a)p:::(x a)q[(x c)2+d2]r:::[(x e)2+f2]sy es ,el integrandodelsegundot erminosepuededescomponercomosigue:R(x)Q( x)=A1(x a)p+A2(x a)p 1+:::+Apx a+:::+B1(x b)q+B2(x b)q 1+:::+Bqx b+M1x+N1((x c)2+d2)r 1+:::+Mrx+Nr(x c)2+d2+:::+H1x+K1((x e)2+f2)s+:::+Hsx+Ks(x e)2+f2. Todaslasconstantesseobtienenidenti candocoe a= lnjx aj+ (x a)p=1(1 p)(x a)p 1+ +N(x c)2+d2dx=M2lnj(x c)2+d2j+Mc+Ndarctanx cd+ +N[(x c)2+d2]rdx)LlamemosIr=RMx+N[(x c)2+d2]rdxyJr=Rdx[(x c)2+d2]rdxoperandoseobtieneIr=M2(1 r) 1((x c)2+d2)r 1+ (Mc+N) JrJr=1d2Jr 1+x cd22(1 r)((x c)2+d2)r 1 1d22(1 r)Jr etododeHermiteSiQ(x) = (x a)m:::(x b)n (x c)2+d2 ::: (x e)2+f2 entoncesZR(x)Q(x)dx=U(x)(x a)m 1:::(x b)n 1:::[(x c)2+d2]p 1:::[(x e)2+f2]q 1+KZdxx a+:::+LZdxx b+ZCx+D(x c)2+d2dx+:::+ZEx+F(x e)2+f2dxdondeU(x) expresi one identi candocoe ondefuncionesirracionalesalgebraicasInte gralesdeltipoZR x; ax+bcx+d m1n1;:::; ax+bcx+d msns!dxja;b;c;d2R;ni;mi2Z;ni6= 0ycydnoseanulansimult cambioax+bcx+d=tmsiendomel m nimocom unm x;pax2+bx+c >0!
7 Pax2+bx+c= pa x+ <0!pax2+bx+c= pc+x ;c <0!pax2+bx+c=t (x ) siendo etodoAlem an:RP(x)pax2+bx+cdx=Q(x) pax2+bx+c+KRdxpax2+bx+cDondegradQ(x) =grad(P(x)) 1 cientesseobtienenderivandola expresi one identi candot omicas:Rxm(a+bxn)pdxja;b2R;m;n;p2Q. !x=tqdondeqesel +1n2Z!a+bxn=tqsiendoqel +1n+p2Z!a+bxnxn=tqsiendoqel ondefuncionestrascendentesSiR(u) es unafunci onracionalyu=f(x) es unafunci onqueadmitefunci oninversaconderivadaracional,entonceslai ntegraldeR(f(x))sereducea unaintegralracionalmedianteel cambiof(x) = ondefuncionestrigonom etricasIntegraci ondeRR(senx;cosx)dx: engeneralsehaceel cambiotanx2=tconloquesenx=2dt1+t2;cosx=1 t21+t2; dx=2t1+t2. (senx;cosx) = R(senx; cosx) sehaceel cambiosenx= (senx;cosx) = R( senx;cosx) sehaceel cambiocosx= (senx;cosx) =R( senx; cosx) sehaceel cambiotanx=tIntegralesdeltipoIm;n=Rsenmx n x dxse ;n=senm+1x cosn 1xm+1+n 1m+1Im+2;n 2; m6= ;n=senm+1x cosn 1xm+n+n 1m+nIm;n 2; m+n6= ;n= senm 1x cosn+1xm+1+m 1n+1Im+2;n+2; m6= ;n= senm 1x cosn+1xm+n+m 1m+nIm 2;n; m+n6= ;n= senm+1x cosn+1xn+1+m+n 2n+1Im;n+2; n6= ;n=senm+1x cosn+1xm+1+m+n 2m+1Im+2;n+2; m6= +p(x)y=q(x) =)y=e R(x)dx Zq(x)eRp(x)dx+C y0+y=p(t) =)y=e t= 1 Zt0p(t)et= dt+ onnum etododeRunge-Kuttadecuartoordeny0=f(x;y) !
8 Yi+1=yi+16(k1+ 2k2+ 2k3+k4)hk1=f(xi;yi)k2=f(xi+h=2;yi+k1h=2) k3=f(xi+h=2;yi+k2h=2)k4=f(xi+h;yi+k3h)12
