Рационаллă функцисенчен илнĕ интегралсен йышĕ

Аяларах рационаллă функцисен интегралĕсĕн (функцисен умсăнарĕсен) йышне паратпăр.

${\displaystyle \int (ax+b)^{n}dx={\begin{cases}{\frac {(ax+b)^{n+1}}{a(n+1)}},&n\neq -1\\{\frac {1}{a}}\ln \left|ax+b\right|,&n=-1\end{cases}}}$

${\displaystyle \int x(ax+b)^{n}dx={\begin{cases}{\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1},&n\not \in \{-1,-2\}\\{\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|,&n=-1\\{\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|,&n=-2\end{cases}}}$

${\displaystyle \int {\frac {x}{(ax+b)^{n}}}dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}},\quad n\not \in \{1,2\}}$

${\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin \left({\frac {(2k-1)\pi }{2^{n}}}\right)\operatorname {arctg} \left[\left(x-\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right)\operatorname {cosec} \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right]\right]-\right.}$ ${\displaystyle \quad \left.-\,{\frac {1}{2^{n}}}\left[\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\ln \left|x^{2}-2x\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)+1\right|\right]\right\}+C}$

${\displaystyle \int {\frac {x^{2}}{ax+b}}dx={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)}$

${\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}dx={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)}$

${\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)}$

${\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx={\frac {1}{a^{3}}}\left(-{\frac {1}{(n-3)(ax+b)^{n-3}}}+{\frac {2b}{(n-2)(ax+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right),}$ для ${\displaystyle n\not \in \{1,2,3\}}$

${\displaystyle \int {\frac {dx}{x(ax+b)}}=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|}$

${\displaystyle \int {\frac {dx}{x^{2}(ax+b)}}=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|}$

${\displaystyle \int {\frac {dx}{x^{2}(ax+b)^{2}}}=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)}$

${\displaystyle \int {\frac {dx}{a^{2}x^{2}+b^{2}}}={\frac {1}{ab}}\operatorname {arctg} {\frac {ax}{b}}}$

${\displaystyle \int {\frac {dx}{(x^{2}+a^{2})^{2}}}={\frac {x}{2a^{2}(x^{2}+a^{2})}}+{\frac {1}{2a^{3}}}\operatorname {arctg} {\frac {x}{a}}}$

${\displaystyle \int {\frac {dx}{(x^{2}+a^{2})^{3}}}={\frac {x}{4a^{2}(x^{2}+a^{2})^{2}}}+{\frac {3x}{8a^{4}(x^{2}+a^{2})}}+{\frac {3}{8a^{5}}}\operatorname {arctg} {\frac {x}{a}}}$

${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\operatorname {arth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}},}$ для ${\displaystyle |x|<|a|}$

${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\operatorname {arcth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}},}$ для ${\displaystyle |x|>|a|}$

${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {4ac-b^{2}}}}\operatorname {arctg} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}},}$ для ${\displaystyle 4ac-b^{2}>0}$

${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}=-{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {arth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|,}$ для ${\displaystyle 4ac-b^{2}<0}$

${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}=-{\frac {2}{2ax+b}}\qquad {\mbox{(for }}4ac-b^{2}=0{\mbox{)}}}$

${\displaystyle \int {\frac {x}{ax^{2}+bx+c}}dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}}$

${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\operatorname {arctg} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}},}$ для ${\displaystyle 4ac-b^{2}>0}$

${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {arth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}},}$ для ${\displaystyle 4ac-b^{2}<0}$

${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}},}$ для ${\displaystyle 4ac-b^{2}=0}$

${\displaystyle \int {\frac {dx}{(ax^{2}+bx+c)^{n}}}={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {dx}{(ax^{2}+bx+c)^{n-1}}}}$

${\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx={\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {dx}{(ax^{2}+bx+c)^{n-1}}}}$

${\displaystyle \int {\frac {dx}{x(ax^{2}+bx+c)}}={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {dx}{ax^{2}+bx+c}}}$