Automorphic number

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The following is a list of integrals (antiderivative functions) of rational functions. For other types of functions, see lists of integrals.


Miscellaneous integrands

f(x)f(x)dx=ln|f(x)|+C
1x2+a2dx=1aarctanxa+C
1x2a2dx={1aarctanhxa=12alnaxa+x+C(for |x|<|a|)1aarccothxa=12alnxax+a+C(for |x|>|a|)
dxx2n+1=k=12n1{12n1[sin((2k1)π2n)arctan[(xcos((2k1)π2n))csc((2k1)π2n)]]12n[cos((2k1)π2n)ln|x22xcos((2k1)π2n)+1|]}+C


Any rational function can be integrated using partial fractions in integration, by decomposing the rational function into a sum of functions of the form:

a(xb)n, and ax+b((xc)2+d2)n.

Integrands of the form xm(a x + b)n

1ax+bdx=1aln|ax+b|+C
More generally,[1]
1ax+bdx={1aln|ax+b|+Cx<b/a1aln|ax+b|+C+x>b/a
(ax+b)ndx=(ax+b)n+1a(n+1)+C(for n1) (Cavalieri's quadrature formula)


xax+bdx=xaba2ln|ax+b|+C
x(ax+b)2dx=ba2(ax+b)+1a2ln|ax+b|+C
x(ax+b)ndx=a(1n)xba2(n1)(n2)(ax+b)n1+C(for n∉{1,2})
x(ax+b)ndx=a(n+1)xba2(n+1)(n+2)(ax+b)n+1+C(for n∉{1,2})


x2ax+bdx=b2ln(|ax+b|)a3+ax22bx2a2+C
x2(ax+b)2dx=1a3(ax2bln|ax+b|b2ax+b)+C
x2(ax+b)3dx=1a3(ln|ax+b|+2bax+bb22(ax+b)2)+C
x2(ax+b)ndx=1a3((ax+b)3n(n3)+2b(ax+b)2n(n2)b2(ax+b)1n(n1))+C(for n∉{1,2,3})


1x(ax+b)dx=1bln|ax+bx|+C
1x2(ax+b)dx=1bx+ab2ln|ax+bx|+C
1x2(ax+b)2dx=a(1b2(ax+b)+1ab2x2b3ln|ax+bx|)+C

Integrands of the form xm / (a x2 + b x + c)n

For a0:

1ax2+bx+cdx={24acb2arctan2ax+b4acb2+C(for 4acb2>0)2b24acarctanh2ax+bb24ac+C=1b24acln|2ax+bb24ac2ax+b+b24ac|+C(for 4acb2<0)22ax+b+C(for 4acb2=0)


xax2+bx+cdx=12aln|ax2+bx+c|b2adxax2+bx+c+C


mx+nax2+bx+cdx={m2aln|ax2+bx+c|+2anbma4acb2arctan2ax+b4acb2+C(for 4acb2>0)m2aln|ax2+bx+c|2anbmab24acarctanh2ax+bb24ac+C(for 4acb2<0)m2aln|ax2+bx+c|2anbma(2ax+b)+C(for 4acb2=0)


1(ax2+bx+c)ndx=2ax+b(n1)(4acb2)(ax2+bx+c)n1+(2n3)2a(n1)(4acb2)1(ax2+bx+c)n1dx+C


x(ax2+bx+c)ndx=bx+2c(n1)(4acb2)(ax2+bx+c)n1b(2n3)(n1)(4acb2)1(ax2+bx+c)n1dx+C


1x(ax2+bx+c)dx=12cln|x2ax2+bx+c|b2c1ax2+bx+cdx+C

Integrands of the form xm (a + b xn)p

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
xm(a+bxn)pdx=xm+1(a+bxn)pm+np+1+anpm+np+1xm(a+bxn)p1dx
xm(a+bxn)pdx=xm+1(a+bxn)p+1an(p+1)+m+n(p+1)+1an(p+1)xm(a+bxn)p+1dx
xm(a+bxn)pdx=xm+1(a+bxn)pm+1bnpm+1xm+n(a+bxn)p1dx
xm(a+bxn)pdx=xmn+1(a+bxn)p+1bn(p+1)mn+1bn(p+1)xmn(a+bxn)p+1dx
xm(a+bxn)pdx=xmn+1(a+bxn)p+1b(m+np+1)a(mn+1)b(m+np+1)xmn(a+bxn)pdx
xm(a+bxn)pdx=xm+1(a+bxn)p+1a(m+1)b(m+n(p+1)+1)a(m+1)xm+n(a+bxn)pdx

Integrands of the form (A + B x) (a + b x)m (c + d x)n (e + f x)p

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, n and p toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
  • Special cases of these reductions formulas can be used for integrands of the form (a+bx)m(c+dx)n(e+fx)p by setting B to 0.
(A+Bx)(a+bx)m(c+dx)n(e+fx)pdx=(AbaB)(a+bx)m+1(c+dx)n(e+fx)p+1b(m+1)(afbe)+1b(m+1)(afbe)
(bc(m+1)(AfBe)+(AbaB)(nde+cf(p+1))+d(b(m+1)(AfBe)+f(n+p+1)(AbaB))x)(a+bx)m+1(c+dx)n1(e+fx)pdx
(A+Bx)(a+bx)m(c+dx)n(e+fx)pdx=B(a+bx)m(c+dx)n+1(e+fx)p+1df(m+n+p+2)+1df(m+n+p+2)
(Aadf(m+n+p+2)B(bcem+a(de(n+1)+cf(p+1)))+(Abdf(m+n+p+2)+B(adfmb(de(m+n+1)+cf(m+p+1))))x)(a+bx)m1(c+dx)n(e+fx)pdx
(A+Bx)(a+bx)m(c+dx)n(e+fx)pdx=(AbaB)(a+bx)m+1(c+dx)n+1(e+fx)p+1(m+1)(adbc)(afbe)+1(m+1)(adbc)(afbe)
((m+1)(A(adfb(cf+de))+Bbce)(AbaB)(de(n+1)+cf(p+1))df(m+n+p+3)(AbaB)x)(a+bx)m+1(c+dx)n(e+fx)pdx

Integrands of the form xm (A + B xn) (a + b xn)p (c + d xn)q

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, p and q toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
  • Special cases of these reductions formulas can be used for integrands of the form (a+bxn)p(c+dxn)q and xm(a+bxn)p(c+dxn)q by setting m and/or B to 0.
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=(AbaB)xm+1(a+bxn)p+1(c+dxn)qabn(p+1)+1abn(p+1)
xm(c(Abn(p+1)+(AbaB)(m+1))+d(Abn(p+1)+(AbaB)(m+nq+1))xn)(a+bxn)p+1(c+dxn)q1dx
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=Bxm+1(a+bxn)p+1(c+dxn)qb(m+n(p+q+1)+1)+1b(m+n(p+q+1)+1)
xm(c((AbaB)(1+m)+Abn(1+p+q))+(d(AbaB)(1+m)+Bnq(bcad)+Abdn(1+p+q))xn)(a+bxn)p(c+dxn)q1dx
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=(AbaB)xm+1(a+bxn)p+1(c+dxn)q+1an(bcad)(p+1)+1an(bcad)(p+1)
xm(c(AbaB)(m+1)+An(bcad)(p+1)+d(AbaB)(m+n(p+q+2)+1)xn)(a+bxn)p+1(c+dxn)qdx
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=Bxmn+1(a+bxn)p+1(c+dxn)q+1bd(m+n(p+q+1)+1)1bd(m+n(p+q+1)+1)
xmn(aBc(mn+1)+(aBd(m+nq+1)b(Bc(m+np+1)+Ad(m+n(p+q+1)+1)))xn)(a+bxn)p(c+dxn)qdx
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=Axm+1(a+bxn)p+1(c+dxn)q+1ac(m+1)+1ac(m+1)
xm+n(aBc(m+1)A(bc+ad)(m+n+1)An(bcp+adq)Abd(m+n(p+q+2)+1)xn)(a+bxn)p(c+dxn)qdx
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=Axm+1(a+bxn)p+1(c+dxn)qa(m+1)1a(m+1)
xm+n(c(AbaB)(m+1)+An(bc(p+1)+adq)+d((AbaB)(m+1)+Abn(p+q+1))xn)(a+bxn)p(c+dxn)q1dx
xm(A+Bxn)(a+bxn)p(c+dxn)qdx=(AbaB)xmn+1(a+bxn)p+1(c+dxn)q+1bn(bcad)(p+1)1bn(bcad)(p+1)
xmn(c(AbaB)(mn+1)+(d(AbaB)(m+nq+1)bn(BcAd)(p+1))xn)(a+bxn)p+1(c+dxn)qdx

Integrands of the form (d + e x)m (a + b x + c x2)p when b2 − 4 a c = 0

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
  • Special cases of these reductions formulas can be used for integrands of the form (a+bx+cx2)p when b24ac=0 by setting m to 0.
(d+ex)m(a+bx+cx2)pdx=(d+ex)m+1(a+bx+cx2)pe(m+1)p(d+ex)m+2(b+2cx)(a+bx+cx2)p1e2(m+1)(m+2p+1)+p(2p1)(2cdbe)e2(m+1)(m+2p+1)(d+ex)m+1(a+bx+cx2)p1dx
(d+ex)m(a+bx+cx2)pdx=(d+ex)m+1(a+bx+cx2)pe(m+1)p(d+ex)m+2(b+2cx)(a+bx+cx2)p1e2(m+1)(m+2)+2cp(2p1)e2(m+1)(m+2)(d+ex)m+2(a+bx+cx2)p1dx
(d+ex)m(a+bx+cx2)pdx=e(m+2p+2)(d+ex)m(a+bx+cx2)p+1(p+1)(2p+1)(2cdbe)+(d+ex)m+1(b+2cx)(a+bx+cx2)p(2p+1)(2cdbe)+e2m(m+2p+2)(p+1)(2p+1)(2cdbe)(d+ex)m1(a+bx+cx2)p+1dx
(d+ex)m(a+bx+cx2)pdx=em(d+ex)m1(a+bx+cx2)p+12c(p+1)(2p+1)+(d+ex)m(b+2cx)(a+bx+cx2)p2c(2p+1)+e2m(m1)2c(p+1)(2p+1)(d+ex)m2(a+bx+cx2)p+1dx
(d+ex)m(a+bx+cx2)pdx=(d+ex)m+1(a+bx+cx2)pe(m+2p+1)p(2cdbe)(d+ex)m+1(b+2cx)(a+bx+cx2)p12ce2(m+2p)(m+2p+1)+p(2p1)(2cdbe)22ce2(m+2p)(m+2p+1)(d+ex)m(a+bx+cx2)p1dx
(d+ex)m(a+bx+cx2)pdx=2ce(m+2p+2)(d+ex)m+1(a+bx+cx2)p+1(p+1)(2p+1)(2cdbe)2+(d+ex)m+1(b+2cx)(a+bx+cx2)p(2p+1)(2cdbe)+2ce2(m+2p+2)(m+2p+3)(p+1)(2p+1)(2cdbe)2(d+ex)m(a+bx+cx2)p+1dx
(d+ex)m(a+bx+cx2)pdx=(d+ex)m(b+2cx)(a+bx+cx2)p2c(m+2p+1)+m(2cdbe)2c(m+2p+1)(d+ex)m1(a+bx+cx2)pdx
(d+ex)m(a+bx+cx2)pdx=(d+ex)m+1(b+2cx)(a+bx+cx2)p(m+1)(2cdbe)+2c(m+2p+2)(m+1)(2cdbe)(d+ex)m+1(a+bx+cx2)pdx

Integrands of the form (d + e x)m (A + B x) (a + b x + c x2)p

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
  • Special cases of these reductions formulas can be used for integrands of the form (a+bx+cx2)p and (d+ex)m(a+bx+cx2)p by setting m and/or B to 0.
(d+ex)m(A+Bx)(a+bx+cx2)pdx=(d+ex)m+1(Ae(m+2p+2)Bd(2p+1)+eB(m+1)x)(a+bx+cx2)pe2(m+1)(m+2p+2)+1e2(m+1)(m+2p+2)p
(d+ex)m+1(B(bd+2ae+2aem+2bdp)Abe(m+2p+2)+(B(2cd+be+bem+4cdp)2Ace(m+2p+2))x)(a+bx+cx2)p1dx
(d+ex)m(A+Bx)(a+bx+cx2)pdx=(d+ex)m(Ab2aB(bB2Ac)x)(a+bx+cx2)p+1(p+1)(b24ac)+1(p+1)(b24ac)
(d+ex)m1(B(2aem+bd(2p+3))A(bem+2cd(2p+3))+e(bB2Ac)(m+2p+3)x)(a+bx+cx2)p+1dx
(d+ex)m(A+Bx)(a+bx+cx2)pdx=(d+ex)m+1(Ace(m+2p+2)B(cd+2cdpbep)+Bce(m+2p+1)x)(a+bx+cx2)pce2(m+2p+1)(m+2p+2)pce2(m+2p+1)(m+2p+2)
(d+ex)m(Ace(bd2ae)(m+2p+2)+B(ae(be2cdm+bem)+bd(bepcd2cdp))+
(Ace(2cdbe)(m+2p+2)B(b2e2(m+p+1)+2c2d2(1+2p)+ce(bd(m2p)+2ae(m+2p+1))))x)(a+bx+cx2)p1dx
(d+ex)m(A+Bx)(a+bx+cx2)pdx=(d+ex)m+1(A(bcdb2e+2ace)aB(2cdbe)+c(A(2cdbe)B(bd2ae))x)(a+bx+cx2)p+1(p+1)(b24ac)(cd2bde+ae2)+
1(p+1)(b24ac)(cd2bde+ae2)
(d+ex)m(A(bcde(2pm+2)+b2e2(m+p+2)2c2d2(3+2p)2ace2(m+2p+3))
B(ae(be2cdm+bem)+bd(3cd+be2cdp+bep))+ce(B(bd2ae)A(2cdbe))(m+2p+4)x)(a+bx+cx2)p+1dx
(d+ex)m(A+Bx)(a+bx+cx2)pdx=B(d+ex)m(a+bx+cx2)p+1c(m+2p+2)+1c(m+2p+2)
(d+ex)m1(m(AcdaBe)d(bB2Ac)(p+1)+((BcdbBe+Ace)me(bB2Ac)(p+1))x)(a+bx+cx2)pdx
(d+ex)m(A+Bx)(a+bx+cx2)pdx=(BdAe)(d+ex)m+1(a+bx+cx2)p+1(m+1)(cd2bde+ae2)+1(m+1)(cd2bde+ae2)
(d+ex)m+1((AcdAbe+aBe)(m+1)+b(BdAe)(p+1)+c(BdAe)(m+2p+3)x)(a+bx+cx2)pdx

Integrands of the form xm (a + b xn + c x2n)p when b2 − 4 a c = 0

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
  • Special cases of these reductions formulas can be used for integrands of the form (a+bxn+cx2n)p when b24ac=0 by setting m to 0.
xm(a+bxn+cx2n)pdx=xm+1(a+bxn+cx2n)pm+2np+1+npxm+1(2a+bxn)(a+bxn+cx2n)p1(m+1)(m+2np+1)bn2p(2p1)(m+1)(m+2np+1)xm+n(a+bxn+cx2n)p1dx
xm(a+bxn+cx2n)pdx=(m+n(2p1)+1)xm+1(a+bxn+cx2n)p(m+1)(m+n+1)+npxm+1(2a+bxn)(a+bxn+cx2n)p1(m+1)(m+n+1)+2cpn2(2p1)(m+1)(m+n+1)xm+2n(a+bxn+cx2n)p1dx
xm(a+bxn+cx2n)pdx=(m+n(2p+1)+1)xmn+1(a+bxn+cx2n)p+1bn2(p+1)(2p+1)xm+1(b+2cxn)(a+bxn+cx2n)pbn(2p+1)(mn+1)(m+n(2p+1)+1)bn2(p+1)(2p+1)xmn(a+bxn+cx2n)p+1dx
xm(a+bxn+cx2n)pdx=(m3n2np+1)xm2n+1(a+bxn+cx2n)p+12cn2(p+1)(2p+1)xm2n+1(2a+bxn)(a+bxn+cx2n)p2cn(2p+1)+(mn+1)(m2n+1)2cn2(p+1)(2p+1)xm2n(a+bxn+cx2n)p+1dx
xm(a+bxn+cx2n)pdx=xm+1(a+bxn+cx2n)pm+2np+1+npxm+1(2a+bxn)(a+bxn+cx2n)p1(m+2np+1)(m+n(2p1)+1)+2an2p(2p1)(m+2np+1)(m+n(2p1)+1)xm(a+bxn+cx2n)p1dx
xm(a+bxn+cx2n)pdx=(m+n+2np+1)xm+1(a+bxn+cx2n)p+12an2(p+1)(2p+1)xm+1(2a+bxn)(a+bxn+cx2n)p2an(2p+1)+(m+n(2p+1)+1)(m+2n(p+1)+1)2an2(p+1)(2p+1)xm(a+bxn+cx2n)p+1dx
xm(a+bxn+cx2n)pdx=xmn+1(b+2cxn)(a+bxn+cx2n)p2c(m+2np+1)b(mn+1)2c(m+2np+1)xmn(a+bxn+cx2n)pdx
xm(a+bxn+cx2n)pdx=xm+1(b+2cxn)(a+bxn+cx2n)pb(m+1)2c(m+n(2p+1)+1)b(m+1)xm+n(a+bxn+cx2n)pdx

Integrands of the form xm (A + B xn) (a + b xn + c x2n)p

  • The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
  • These reduction formulas can be used for integrands having integer and/or fractional exponents.
  • Special cases of these reductions formulas can be used for integrands of the form (a+bxn+cx2n)p and xm(a+bxn+cx2n)p by setting m and/or B to 0.
xm(A+Bxn)(a+bxn+cx2n)pdx=xm+1(A(m+n(2p+1)+1)+B(m+1)xn)(a+bxn+cx2n)p(m+1)(m+n(2p+1)+1)+np(m+1)(m+n(2p+1)+1)
xm+n(2aB(m+1)Ab(m+n(2p+1)+1)+(bB(m+1)2Ac(m+n(2p+1)+1))xn)(a+bxn+cx2n)p1dx
xm(A+Bxn)(a+bxn+cx2n)pdx=xmn+1(Ab2aB(bB2Ac)xn)(a+bxn+cx2n)p+1n(p+1)(b24ac)+1n(p+1)(b24ac)
xmn((mn+1)(2aBAb)+(m+2n(p+1)+1)(bB2Ac)xn)(a+bxn+cx2n)p+1dx
xm(A+Bxn)(a+bxn+cx2n)pdx=xm+1(bBnp+Ac(m+n(2p+1)+1)+Bc(m+2np+1)xn)(a+bxn+cx2n)pc(m+2np+1)(m+n(2p+1)+1)+npc(m+2np+1)(m+n(2p+1)+1)
xm(2aAc(m+n(2p+1)+1)abB(m+1)+(2aBc(m+2np+1)+Abc(m+n(2p+1)+1)b2B(m+np+1))xn)(a+bxn+cx2n)p1dx
xm(A+Bxn)(a+bxn+cx2n)pdx=xm+1(Ab2abB2aAc+(Ab2aB)cxn)(a+bxn+cx2n)p+1an(p+1)(b24ac)+1an(p+1)(b24ac)
xm((m+n(p+1)+1)Ab2abB(m+1)2(m+2n(p+1)+1)aAc+(m+n(2p+3)+1)(Ab2aB)cxn)(a+bxn+cx2n)p+1dx
xm(A+Bxn)(a+bxn+cx2n)pdx=Bxmn+1(a+bxn+cx2n)p+1c(m+n(2p+1)+1)1c(m+n(2p+1)+1)
xmn(aB(mn+1)+(bB(m+np+1)Ac(m+n(2p+1)+1))xn)(a+bxn+cx2n)pdx
xm(A+Bxn)(a+bxn+cx2n)pdx=Axm+1(a+bxn+cx2n)p+1a(m+1)+1a(m+1)
xm+n(aB(m+1)Ab(m+n(p+1)+1)Ac(m+2n(p+1)+1)xn)(a+bxn+cx2n)pdx

References

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Template:Lists of integrals

  1. "Reader Survey: log|x| + C", Tom Leinster, The n-category Café, March 19, 2012