Daftar integral dari fungsi rasional

tabel integral.
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 \int (ax+b)^{n}dx:  ={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{(untuk }}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{(untuk }}n\not \in \{-1,-2\}{\mbox{)}}
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 \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{(untuk }}n\not \in \{1,2\}{\mbox{)}}
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 \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{(untuk }}n\not \in \{1,2,3\}{\mbox{)}}
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 \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{(untuk }}|x|<|a|{\mbox{)}}\,\!
:   -{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{(untuk }}|x|>|a|{\mbox{)}}\,\!
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 \int {\frac {1}{ax^{2}+bx+c}}dx=:   {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(untuk }}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{(untuk }}4ac-b^{2}<0{\mbox{)}}
:   -{\frac {2}{2ax+b}}\qquad {\mbox{(untuk }}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}}
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 \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{(untuk }}4ac-b^{2}>0{\mbox{)}}
:   {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(untuk }}4ac-b^{2}<0{\mbox{)}}
:   {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}\,\,\,\,\,\,\,\,\,\,\qquad {\mbox{(untuk }}4ac-b^{2}=0{\mbox{)}}
Fungsi rasional apapun dapat diintegrasikan melalui persamaan-persamaan di atas dengan memanfaatkan integrasi parsial, dengan menguraikan fungsi rasional menjadi penjumlahan fungsi-fungsi dalam bentuk