Całki funkcji wymiernych

Lista całek nieoznaczonych funkcji wymiernych

Lista całek nieoznaczonych funkcji wymiernych

Lista innych całek znajduje się w tablicy całek. Algorytm całkowania dowolnej funkcji wymiernej jest podany w tym artykule.

Całki funkcji

wymiernych | logarytmicznych | wykładniczych | niewymiernych | trygonometrycznych | hiperbolicznych | arcus | area

$\int 1\;{\mbox{d}}x=x$

$\int x^{n}{\mbox{d}}x={\frac {x^{n+1}}{n+1}}\qquad (n\neq -1)$

$\int (ax+b)^{n}{\mbox{d}}x={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{(dla }}n\neq -1{\mbox{)}}$

$\int {\frac {1}{x}}{\mbox{d}}x=\ln |x|$

$\int {\frac {1}{x^{n}}}{\mbox{d}}x={\frac {-1}{(n-1)x^{n-1}}}\qquad (n\neq 1)$

$\int {\frac {{\mbox{d}}x}{ax+b}}={\frac {1}{a}}\ln \left|ax+b\right|$

$\int x(ax+b)^{n}{\mbox{d}}x={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{(dla }}n\not \in \{-1,-2\}{\mbox{)}}$

$\int {\frac {x\;{\mbox{d}}x}{ax+b}}={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|$

$\int {\frac {x\;{\mbox{d}}x}{(ax+b)^{2}}}={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|$

$\int {\frac {x\;{\mbox{d}}x}{(ax+b)^{n}}}={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}\qquad {\mbox{(dla }}n\not \in \{1,2\}{\mbox{)}}$

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

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

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

$\int {\frac {x^{2}\;{\mbox{d}}x}{(ax+b)^{n}}}={\frac {1}{a^{3}}}\left(-{\frac {1}{(n-3)(ax+b)^{n-3}}}+{\frac {2b}{(n-2)(a+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right)\qquad {\mbox{(dla }}n\not \in \{1,2,3\}{\mbox{)}}$

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

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

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

$\int {\frac {{\mbox{d}}x}{x^{2}+a^{2}}}={\frac {1}{a}}\mathrm {arctg} {\frac {x}{a}}$

$\int {\frac {{\mbox{d}}x}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arctgh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{(dla }}|x|>|a|{\mbox{)}}$

$\int {\frac {{\mbox{d}}x}{ax^{2}+bx+c}}={\frac {2}{\sqrt {4ac-b^{2}}}}\mathrm {arctg} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(dla }}4ac-b^{2}>0{\mbox{)}}$

$\int {\frac {{\mbox{d}}x}{ax^{2}+bx+c}}=-{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {artgh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|\qquad {\mbox{(dla }}4ac-b^{2}<0{\mbox{)}}$

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

$\int {\frac {mx+n}{ax^{2}+bx+c}}{\mbox{d}}x={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\mathrm {arctg} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(dla }}4ac-b^{2}>0{\mbox{)}}$

$\int {\frac {mx+n}{ax^{2}+bx+c}}{\mbox{d}}x={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artgh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(dla }}4ac-b^{2}<0{\mbox{)}}$

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

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

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

$\int {\frac {{\mbox{d}}x}{(a+bx^{n})^{p}}}={\frac {x}{(n(p-1)a)(a+bx^{n})^{p-1}}}-{\frac {1-np+n}{n(p-1)a}}\int {\frac {{\mbox{d}}x}{(a+bx^{n})^{p-1}}}$

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