Ondorengoa funtzio arrazionalen integralen zerrenda bat da (jatorrizkoak edo antideribatuak). Integralen zerrenda osatuago nahi baduzu, ikusi integralen zerrenda.















![{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}dx={\begin{cases}\displaystyle -{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+K&{\mbox{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+K&{\mbox{( }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4211b35a9535ac4ec88df4abdfccc1ccce7f81a)
bada:
![{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+K&{\mbox{( }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+K={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+K&{\mbox{( }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+K&{\mbox{( }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd2863cf805a50629ae0466988426ee30e7cfaaf)

![{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+K&{\mbox{( }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+K&{\mbox{( }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+K&{\mbox{( }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dbff163b1e8ad2bc4b3f41e785c7c38af074502)



![{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin \left({\frac {(2k-1)\pi }{2^{n}}}\right)\arctan \left[\left(x-\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right)\csc \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right]\right]-{\frac {1}{2^{n}}}\left[\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\ln \left|x^{2}-2x\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)+1\right|\right]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/493e98d39eebbb6342aa5c928ba6da54755b63fe)
Edozein funtzio arrazional integra dezakegu goiko berdintzak erabiliz eta zatiki arrazionalen integrazioaren artikuluan eskaintzen diren teknikak aplikatuz, funtzio arrazionalak ondorengo formako batugaietan banatzearen bidez:
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