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== Integral Forms & Antiderivatives ==
== Integral Forms & Antiderivatives ==
=== Basic Forms ===
 
=== Basic Forms: ===
# <math>\int u \, dv = uv - \int v \, du</math><br>
# <math>\int u \, dv = uv - \int v \, du</math><br>
# <math>\int u^n \, du = \frac{u^{n+1}}{n+1} + C, \quad (n \neq -1)</math><br>
# <math>\int u^n \, du = \frac{u^{n+1}}{n+1} + C, \quad (n \neq -1)</math><br>
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# <math>\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C</math><br>
# <math>\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C</math><br>
<br>
<br>
=== Forms Involving <math>\sqrt{a}</math> ===


=== Forms Involving: <math>\sqrt{a^2 + u^2},\ a > 0</math> ===
# <math>\int \sqrt{a^2 + u^2}\, du = \frac{u}{2}\sqrt{a^2 + u^2} + \frac{a^2}{2}\ln(u + \sqrt{a^2 + u^2}) + C</math><br>
# <math>\int u^2\sqrt{a^2 + u^2}\, du = \frac{u}{8}(a^2 + 2u^2)\sqrt{a^2 + u^2} - \frac{a^4}{8}\ln(u + \sqrt{a^2 + u^2}) + C</math><br>
# <math>\int \frac{\sqrt{a^2 + u^2}}{u}\, du = \sqrt{a^2 + u^2} - a\ln\left|\frac{a + \sqrt{a^2 + u^2}}{u}\right| + C</math><br>
# <math>\int \frac{\sqrt{a^2 + u^2}}{u^2}\, du = -\frac{\sqrt{a^2 + u^2}}{u} + \ln(u + \sqrt{a^2 + u^2}) + C</math><br>
# <math>\int \frac{du}{\sqrt{a^2 + u^2}} = \ln(u + \sqrt{a^2 + u^2}) + C</math><br>
# <math>\int \frac{u^2\, du}{\sqrt{a^2 + u^2}} = \frac{u}{2}\sqrt{a^2 + u^2} - \frac{a^2}{2}\ln(u + \sqrt{a^2 + u^2}) + C</math><br>
# <math>\int \frac{du}{u\sqrt{a^2 + u^2}} = -\frac{1}{a}\ln\left|\frac{\sqrt{a^2 + u^2} + a}{u}\right| + C</math><br>
# <math>\int \frac{du}{u^2\sqrt{a^2 + u^2}} = -\frac{\sqrt{a^2 + u^2}}{a^2 u} + C</math><br>
# <math>\int \frac{du}{(a^2 + u^2)^{3/2}} = \frac{u}{a^2\sqrt{a^2 + u^2}} + C</math><br>
<br>
=== Forms Involving: <math>\sqrt{a^2 - u^2},\ a > 0</math> ===
# <math>\int \sqrt{a^2 - u^2}\, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C</math><br>
# <math>\int u^2\sqrt{a^2 - u^2}\, du = \frac{u}{8}(2u^2 - a^2)\sqrt{a^2 - u^2} + \frac{a^4}{8}\sin^{-1}\frac{u}{a} + C</math><br>
# <math>\int \frac{\sqrt{a^2 - u^2}}{u}\, du = \sqrt{a^2 - u^2} - a\ln\left|\frac{a + \sqrt{a^2 - u^2}}{u}\right| + C</math><br>
# <math>\int \frac{\sqrt{a^2 - u^2}}{u^2}\, du = -\frac{1}{u}\sqrt{a^2 - u^2} - \sin^{-1}\frac{u}{a} + C</math><br>
# <math>\int \frac{u^2\, du}{\sqrt{a^2 - u^2}} = -\frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C</math><br>
# <math>\int \frac{du}{u\sqrt{a^2 - u^2}} = -\frac{1}{a}\ln\left|\frac{a + \sqrt{a^2 - u^2}}{u}\right| + C</math><br>
# <math>\int \frac{du}{u^2\sqrt{a^2 - u^2}} = -\frac{1}{a^2 u}\sqrt{a^2 - u^2} + C</math><br>
# <math>\int (a^2 - u^2)^{3/2}\, du = -\frac{u}{8}(2u^2 - 5a^2)\sqrt{a^2 - u^2} + \frac{3a^4}{8}\sin^{-1}\frac{u}{a} + C</math><br>
# <math>\int \frac{du}{(a^2 - u^2)^{3/2}} = \frac{u}{a^2\sqrt{a^2 - u^2}} + C</math><br>
<br>
=== Forms Involving: <math>\sqrt{u^2 - a^2},\ a > 0</math> ===
# <math>\int \sqrt{u^2 - a^2}\, du = \frac{u}{2}\sqrt{u^2 - a^2} - \frac{a^2}{2}\ln\left|u + \sqrt{u^2 - a^2}\right| + C</math><br>
# <math>\int u^2\sqrt{u^2 - a^2}\, du = \frac{u}{8}(2u^2 - a^2)\sqrt{u^2 - a^2} - \frac{a^4}{8}\ln\left|u + \sqrt{u^2 - a^2}\right| + C</math><br>
# <math>\int \frac{\sqrt{u^2 - a^2}}{u}\, du = \sqrt{u^2 - a^2} - a\cos^{-1}\frac{a}{|u|} + C</math><br>
# <math>\int \frac{\sqrt{u^2 - a^2}}{u^2}\, du = -\frac{\sqrt{u^2 - a^2}}{u} + \ln\left|u + \sqrt{u^2 - a^2}\right| + C</math><br>
# <math>\int \frac{du}{\sqrt{u^2 - a^2}} = \ln\left|u + \sqrt{u^2 - a^2}\right| + C</math><br>
# <math>\int \frac{u^2\, du}{\sqrt{u^2 - a^2}} = \frac{u}{2}\sqrt{u^2 - a^2} + \frac{a^2}{2}\ln\left|u + \sqrt{u^2 - a^2}\right| + C</math><br>
# <math>\int \frac{du}{u^2\sqrt{u^2 - a^2}} = \frac{\sqrt{u^2 - a^2}}{a^2 u} + C</math><br>
# <math>\int \frac{du}{(u^2 - a^2)^{3/2}} = -\frac{u}{a^2\sqrt{u^2 - a^2}} + C</math><br>
<br>
=== Forms Involving: <math>a + bu</math> ===
# <math>\int \frac{u\, du}{a + bu} = \frac{1}{b^2}(a + bu - a\ln|a + bu|) + C</math>
# <math>\int \frac{u^2\, du}{a + bu} = \frac{1}{2b^3}\left[(a + bu)^2 - 4a(a + bu) + 2a^2\ln|a + bu|\right] + C</math>
# <math>\int \frac{du}{u(a + bu)} = \frac{1}{a}\ln\left|\frac{u}{a + bu}\right| + C</math>
# <math>\int \frac{du}{u^2(a + bu)} = -\frac{1}{au} + \frac{b}{a^2}\ln\left|\frac{a + bu}{u}\right| + C</math>
# <math>\int \frac{u\, du}{(a + bu)^2} = \frac{a}{b^2(a + bu)} + \frac{1}{b^2}\ln|a + bu| + C</math>
# <math>\int \frac{du}{u(a + bu)^2} = \frac{1}{a(a + bu)} - \frac{1}{a^2}\ln\left|\frac{a + bu}{u}\right| + C</math>
# <math>\int \frac{u^2\, du}{(a + bu)^2} = \frac{1}{b^3}\left(a + bu - \frac{a^2}{a + bu} - 2a\ln|a + bu|\right) + C</math>
# <math>\int u\sqrt{a + bu}\, du = \frac{2}{15b^2}(3bu - 2a)(a + bu)^{3/2} + C</math>
# <math>\int \frac{u\, du}{\sqrt{a + bu}} = \frac{2}{3b^2}(bu - 2a)\sqrt{a + bu} + C</math>
# <math>\int \frac{u^2\, du}{\sqrt{a + bu}} = \frac{2}{15b^3}(8a^2 + 3b^2u^2 - 4abu)\sqrt{a + bu} + C</math>
# <math>\int \frac{du}{u\sqrt{a + bu}} = \begin{cases} \dfrac{1}{\sqrt{a}}\ln\left|\dfrac{\sqrt{a + bu} - \sqrt{a}}{\sqrt{a + bu} + \sqrt{a}}\right| + C & \text{if } a > 0 \\[6pt] \dfrac{2}{\sqrt{-a}}\tan^{-1}\sqrt{\dfrac{a + bu}{-a}} + C & \text{if } a < 0 \end{cases}</math>
# <math>\int \frac{\sqrt{a + bu}}{u}\, du = 2\sqrt{a + bu} + a\int \frac{du}{u\sqrt{a + bu}}</math>
# <math>\int \frac{\sqrt{a + bu}}{u^2}\, du = -\frac{\sqrt{a + bu}}{u} + \frac{b}{2}\int \frac{du}{u\sqrt{a + bu}}</math>
# <math>\int u^n\sqrt{a + bu}\, du = \frac{2}{b(2n + 3)}\left[u^n(a + bu)^{3/2} - na\int u^{n-1}\sqrt{a + bu}\, du\right]</math>
# <math>\int \frac{u^n\, du}{\sqrt{a + bu}} = \frac{2u^n\sqrt{a + bu}}{b(2n + 1)} - \frac{2na}{b(2n + 1)}\int \frac{u^{n-1}\, du}{\sqrt{a + bu}}</math>
# <math>\int \frac{du}{u^n\sqrt{a + bu}} = -\frac{\sqrt{a + bu}}{a(n-1)u^{n-1}} - \frac{b(2n-3)}{2a(n-1)}\int \frac{du}{u^{n-1}\sqrt{a + bu}}</math>
<br>
== To Do ==
== To Do ==
*Antiderivative DB search system.
*Antiderivative DB search system.
*Find a use for the white-space on the right of the page.
*Find a use for the white-space on the right of the page.
*Integral Isolating System for easy reference/copying.

Revision as of 15:45, 24 May 2026

Integral Forms & Antiderivatives

Basic Forms:

  1. udv=uvvdu
  2. undu=un+1n+1+C,(n1)
  3. duu=ln|u|+C
  4. eudu=eu+C
  5. audu=aulna+C
  6. sinudu=cosu+C
  7. cosudu=sinu+C
  8. sec2udu=tanu+C
  9. csc2udu=cotu+C
  10. secutanudu=secu+C
  11. cscucotudu=cscu+C
  12. tanudu=ln|secu|+C
  13. cotudu=ln|sinu|+C
  14. secudu=ln|secu+tanu|+C
  15. cscudu=ln|cscucotu|+C
  16. dua2u2=sin1ua+C,(a>0)
  17. dua2+u2=1atan1ua+C
  18. duuu2a2=1asec1ua+C
  19. dua2u2=12aln|u+aua|+C
  20. duu2a2=12aln|uau+a|+C


Forms Involving: a2+u2, a>0

  1. a2+u2du=u2a2+u2+a22ln(u+a2+u2)+C
  2. u2a2+u2du=u8(a2+2u2)a2+u2a48ln(u+a2+u2)+C
  3. a2+u2udu=a2+u2aln|a+a2+u2u|+C
  4. a2+u2u2du=a2+u2u+ln(u+a2+u2)+C
  5. dua2+u2=ln(u+a2+u2)+C
  6. u2dua2+u2=u2a2+u2a22ln(u+a2+u2)+C
  7. duua2+u2=1aln|a2+u2+au|+C
  8. duu2a2+u2=a2+u2a2u+C
  9. du(a2+u2)3/2=ua2a2+u2+C


Forms Involving: a2u2, a>0

  1. a2u2du=u2a2u2+a22sin1ua+C
  2. u2a2u2du=u8(2u2a2)a2u2+a48sin1ua+C
  3. a2u2udu=a2u2aln|a+a2u2u|+C
  4. a2u2u2du=1ua2u2sin1ua+C
  5. u2dua2u2=u2a2u2+a22sin1ua+C
  6. duua2u2=1aln|a+a2u2u|+C
  7. duu2a2u2=1a2ua2u2+C
  8. (a2u2)3/2du=u8(2u25a2)a2u2+3a48sin1ua+C
  9. du(a2u2)3/2=ua2a2u2+C


Forms Involving: u2a2, a>0

  1. u2a2du=u2u2a2a22ln|u+u2a2|+C
  2. u2u2a2du=u8(2u2a2)u2a2a48ln|u+u2a2|+C
  3. u2a2udu=u2a2acos1a|u|+C
  4. u2a2u2du=u2a2u+ln|u+u2a2|+C
  5. duu2a2=ln|u+u2a2|+C
  6. u2duu2a2=u2u2a2+a22ln|u+u2a2|+C
  7. duu2u2a2=u2a2a2u+C
  8. du(u2a2)3/2=ua2u2a2+C


Forms Involving: a+bu

  1. udua+bu=1b2(a+bualn|a+bu|)+C
  2. u2dua+bu=12b3[(a+bu)24a(a+bu)+2a2ln|a+bu|]+C
  3. duu(a+bu)=1aln|ua+bu|+C
  4. duu2(a+bu)=1au+ba2ln|a+buu|+C
  5. udu(a+bu)2=ab2(a+bu)+1b2ln|a+bu|+C
  6. duu(a+bu)2=1a(a+bu)1a2ln|a+buu|+C
  7. u2du(a+bu)2=1b3(a+bua2a+bu2aln|a+bu|)+C
  8. ua+budu=215b2(3bu2a)(a+bu)3/2+C
  9. udua+bu=23b2(bu2a)a+bu+C
  10. u2dua+bu=215b3(8a2+3b2u24abu)a+bu+C
  11. duua+bu={1aln|a+buaa+bu+a|+Cif a>02atan1a+bua+Cif a<0
  12. a+buudu=2a+bu+aduua+bu
  13. a+buu2du=a+buu+b2duua+bu
  14. una+budu=2b(2n+3)[un(a+bu)3/2naun1a+budu]
  15. undua+bu=2una+bub(2n+1)2nab(2n+1)un1dua+bu
  16. duuna+bu=a+bua(n1)un1b(2n3)2a(n1)duun1a+bu


To Do

  • Antiderivative DB search system.
  • Find a use for the white-space on the right of the page.
  • Integral Isolating System for easy reference/copying.