# Llista d'integrals de funcions racionals

Qualsevol fracció racional es pot integrar (trobar la seva primitiva) emprant les identitats que es presenten en aquesta pàgina i les tècniques que es descriuen a "integració de fraccions racionals" a base de descompondre la fracció racional en la suma de funcions de la forma:
    ${\frac {ex+f}{\left(ax^{2}+bx+c\right)^{n}}}$.
A totes les funcions resultat de la integració cal afegir-hi una constant d'integració arbitrària que no s'ha posat a la taula.
    $\int (ax+b)^{n}dx$$={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{(per }}n\neq -1{\mbox}}\,\!$ $\int {\frac {1}{ax+b}}dx$$={\frac {1}{a}}\ln \left|ax+b\right|$ $\int x(ax+b)^{n}dx$$={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{(per }}n\not \in \{-1,-2\}{\mbox}}$
    $\int {\frac {x}{ax+b}}dx$$={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|$ $\int {\frac {x}{(ax+b)^{2}}}dx$$={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|$ $\int {\frac {x}{(ax+b)^{n}}}dx$$={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}\qquad {\mbox{(per }}n\not \in \{1,2\}{\mbox}}$
    $\int {\frac {x^{2}}{ax+b}}dx$$={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)$ $\int {\frac {x^{2}}{(ax+b)^{2}}}dx$$={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)$ $\int {\frac {x^{2}}{(ax+b)^{3}}}dx$$={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)$ $\int {\frac {x^{2}}{(ax+b)^{n}}}dx$$={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(a+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)\qquad {\mbox{(per }}n\not \in \{1,2,3\}{\mbox}}$
    $\int {\frac {1}{x(ax+b)}}dx$$=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|$ $\int {\frac {1}{x^{2}(ax+b)}}dx$$=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|$ $\int {\frac {1}{x^{2}(ax+b)^{2}}}dx$$=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)$ $\int {\frac {1}{x^{2}+a^{2}}}dx$$={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!$ $\int {\frac {1}{x^{2}-a^{2}}}dx=$ $-{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}\qquad {\mbox{(per }}|x|<|a|{\mbox}}\,\!$ $-{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{(per }}|x|>|a|{\mbox}}\,\!$
per $a\neq 0:$
    $\int {\frac {1}{ax^{2}+bx+c}}dx=$ ${\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(per }}4ac-b^{2}>0{\mbox}}$ $-{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|\qquad {\mbox{(per }}4ac-b^{2}<0{\mbox}}$ $-{\frac {2}{2ax+b}}\qquad {\mbox{(per }}4ac-b^{2}=0{\mbox}}$ $\int {\frac {x}{ax^{2}+bx+c}}dx$$={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}$
    $\int {\frac {mx+n}{ax^{2}+bx+c}}dx=$ ${\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(per }}4ac-b^{2}>0{\mbox}}$ ${\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(per }}4ac-b^{2}<0{\mbox}}$ ${\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}\,\,\,\,\,\,\,\,\,\,\qquad {\mbox{(per }}4ac-b^{2}=0{\mbox}}$
    $\int {\frac {1}{(ax^{2}+bx+c)^{n}}}dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx\,\!$
    $\int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx={\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx\,\!$
    $\int {\frac {1}{x(ax^{2}+bx+c)}}dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}dx$
    $\int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin({\frac {(2k-1)\pi }{2^{n}}})\arctan[\left(x-\cos({\frac {(2k-1)\pi }{2^{n}}})\right)\csc({\frac {(2k-1)\pi }{2^{n}}})]\right]-{\frac {1}{2^{n}}}\left[\cos({\frac {(2k-1)\pi }{2^{n}}})\ln \left|x^{2}-2x\cos({\frac {(2k-1)\pi }{2^{n}}})+1\right|\right]\right\}$