# Tích phân hàm số toán cotangent

${\displaystyle \int \cot ax\;dx={\frac {1}{a}}\ln |\sin ax|+C\,\!}$
${\displaystyle \int \cot ^{n}ax\;dx=-{\frac {1}{a(n-1)}}\cot ^{n-1}ax-\int \cot ^{n-2}ax\;dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{1+\cot ax}}=\int {\frac {\tan ax\;dx}{\tan ax+1}}\,\!}$
${\displaystyle \int {\frac {dx}{1-\cot ax}}=\int {\frac {\tan ax\;dx}{\tan ax-1}}\,\!}$

${\displaystyle \int {\frac {dx}{\cos ax\pm \sin ax}}={\frac {1}{a{\sqrt {2}}}}\ln \left|\tan \left({\frac {ax}{2}}\pm {\frac {\pi }{8}}\right)\right|+C}$
${\displaystyle \int {\frac {dx}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C}$
${\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left({\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right)}$
${\displaystyle \int {\frac {\cos ax\;dx}{\cos ax+\sin ax}}={\frac {x}{2}}+{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C}$
${\displaystyle \int {\frac {\cos ax\;dx}{\cos ax-\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C}$
${\displaystyle \int {\frac {\sin ax\;dx}{\cos ax+\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C}$
${\displaystyle \int {\frac {\sin ax\;dx}{\cos ax-\sin ax}}=-{\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C}$
${\displaystyle \int {\frac {\cos ax\;dx}{\sin ax(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}$
${\displaystyle \int {\frac {\cos ax\;dx}{\sin ax(1+-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}$
${\displaystyle \int {\frac {\sin ax\;dx}{\cos ax(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}$
${\displaystyle \int {\frac {\sin ax\;dx}{\cos ax(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}$
${\displaystyle \int \sin ax\cos ax\;dx={\frac {1}{2a}}\sin ^{2}ax+C\,\!}$
${\displaystyle \int \sin a_{1}x\cos a_{2}x\;dx=-{\frac {\cos(a_{1}+a_{2})x}{2(a_{1}+a_{2})}}-{\frac {\cos(a_{1}-a_{2})x}{2(a_{1}-a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!}$
${\displaystyle \int \sin ^{n}ax\cos ax\;dx={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}$
${\displaystyle \int \sin ax\cos ^{n}ax\;dx=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}$
${\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;dx=-{\frac {\sin ^{n-1}ax\cos ^{m+1}ax}{a(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}ax\cos ^{m}ax\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}$
also: ${\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;dx={\frac {\sin ^{n+1}ax\cos ^{m-1}ax}{a(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}ax\cos ^{m-2}ax\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{\sin ax\cos ax}}={\frac {1}{a}}\ln \left|\tan ax\right|+C}$
${\displaystyle \int {\frac {dx}{\sin ax\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {dx}{\sin ax\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{\sin ^{n}ax\cos ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {dx}{\sin ^{n-2}ax\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\sin ax\;dx}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\sin ^{2}ax\;dx}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {ax}{2}}\right)\right|+C}$
${\displaystyle \int {\frac {\sin ^{2}ax\;dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\;dx}{\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}={\frac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}ax\;dx}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$
also: ${\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}=-{\frac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}ax\;dx}{\cos ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}$
also: ${\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}={\frac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}ax\;dx}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\cos ax\;dx}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\cos ^{2}ax\;dx}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left|\tan {\frac {ax}{2}}\right|\right)+C}$
${\displaystyle \int {\frac {\cos ^{2}ax\;dx}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax)}}+\int {\frac {dx}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}=-{\frac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-m-2}{m-1}}\int {\frac {\cos ^{n}ax\;dx}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$
also: ${\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}={\frac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\cos ^{n-2}ax\;dx}{\sin ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}$
also: ${\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}=-{\frac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\cos ^{n-2}ax\;dx}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int \sin ax\tan ax\;dx={\frac {1}{a}}(\ln |\sec ax+\tan ax|-\sin ax)+C\,\!}$
${\displaystyle \int {\frac {\tan ^{n}ax\;dx}{\sin ^{2}ax}}={\frac {1}{a(n-1)}}\tan ^{n-1}(ax)+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\tan ^{n}ax\;dx}{\cos ^{2}ax}}={\frac {1}{a(n+1)}}\tan ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cot ^{n}ax\;dx}{\sin ^{2}ax}}={\frac {1}{a(n+1)}}\cot ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cot ^{n}ax\;dx}{\cos ^{2}ax}}={\frac {1}{a(1-n)}}\tan ^{1-n}ax+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int _{-c}^{c}\sin {x}\;dx=0\!}$
${\displaystyle \int _{-c}^{c}\cos {x}\;dx=2\int _{0}^{c}\cos {x}\;dx=2\int _{-c}^{0}\cos {x}\;dx=2\sin {c}\!}$
${\displaystyle \int _{-c}^{c}\tan {x}\;dx=0\!}$
${\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=1,3,5...{\mbox{)}}\,\!}$

${\displaystyle \int \arcsin {\frac {x}{c}}\,dx=x\arcsin {\frac {x}{c}}+{\sqrt {c^{2}-x^{2}}}}$
${\displaystyle \int x\arcsin {\frac {x}{c}}\,dx=\left({\frac {x^{2}}{2}}-{\frac {c^{2}}{4}}\right)\arcsin {\frac {x}{c}}+{\frac {x}{4}}{\sqrt {c^{2}-x^{2}}}}$
${\displaystyle \int x^{2}\arcsin {\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\arcsin {\frac {x}{c}}+{\frac {x^{2}+2c^{2}}{9}}{\sqrt {c^{2}-x^{2}}}}$
${\displaystyle \int x^{n}\sin ^{-1}x\,dx={\frac {1}{n+1}}\left(x^{n+1}\sin ^{-1}x\right.}$
${\displaystyle \left.+{\frac {x^{n}{\sqrt {1-x^{2}}}-nx^{n-1}\sin ^{-1}x}{n-1}}+n\int x^{n-2}\sin ^{-1}x\,dx\right)}$
${\displaystyle \int \arccos {\frac {x}{c}}\,dx=x\arccos {\frac {x}{c}}-{\sqrt {c^{2}-x^{2}}}}$
${\displaystyle \int x\arccos {\frac {x}{c}}\,dx=\left({\frac {x^{2}}{2}}-{\frac {c^{2}}{4}}\right)\arccos {\frac {x}{c}}-{\frac {x}{4}}{\sqrt {c^{2}-x^{2}}}}$
${\displaystyle \int x^{2}\arccos {\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\arccos {\frac {x}{c}}-{\frac {x^{2}+2c^{2}}{9}}{\sqrt {c^{2}-x^{2}}}}$
${\displaystyle \int \arctan {\frac {x}{c}}\,dx=x\arctan {\frac {x}{c}}-{\frac {c}{2}}\ln(c^{2}+x^{2})}$
${\displaystyle \int x\arctan {\frac {x}{c}}\,dx={\frac {c^{2}+x^{2}}{2}}\arctan {\frac {x}{c}}-{\frac {cx}{2}}}$
${\displaystyle \int x^{2}\arctan {\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\arctan {\frac {x}{c}}-{\frac {cx^{2}}{6}}+{\frac {c^{3}}{6}}\ln {c^{2}+x^{2}}}$
${\displaystyle \int x^{n}\arctan {\frac {x}{c}}\,dx={\frac {x^{n+1}}{n+1}}\arctan {\frac {x}{c}}-{\frac {c}{n+1}}\int {\frac {x^{n+1}dx}{c^{2}+x^{2}}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int \operatorname {arcsec} {\frac {x}{c}}\,dx=x\operatorname {arcsec} {\frac {x}{c}}+{\frac {x}{c|x|}}\ln {|x\pm {\sqrt {x^{2}-1}}|}}$
${\displaystyle \int x\operatorname {arcsec} {x}\,dx\,=\,{\frac {1}{2}}\left(x^{2}\operatorname {arcsec} {x}-{\sqrt {x^{2}-1}}\right)}$
${\displaystyle \int x^{n}\operatorname {arcsec} {x}\,dx\,=\,{\frac {1}{n+1}}\left(x^{n+1}\operatorname {arcsec} {x}-{\frac {1}{n}}\left(x^{n-1}{\sqrt {x^{2}-1}}\;\right.\right.}$
${\displaystyle \left.\left.+(1-n)\left(x^{n-1}\operatorname {arcsec} {x}+(1-n)\int x^{n-2}\operatorname {arcsec} {x}\,dx\right)\right)\right)}$
${\displaystyle \int \mathrm {arccot} \,{\frac {x}{c}}\,dx=x\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {c}{2}}\ln(c^{2}+x^{2})}$
${\displaystyle \int x\,\mathrm {arccot} \,{\frac {x}{c}}\,dx={\frac {c^{2}+x^{2}}{2}}\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {cx}{2}}}$
${\displaystyle \int x^{2}\,\mathrm {arccot} \,{\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {cx^{2}}{6}}-{\frac {c^{3}}{6}}\ln(c^{2}+x^{2})}$
${\displaystyle \int x^{n}\,\mathrm {arccot} \,{\frac {x}{c}}\,dx={\frac {x^{n+1}}{n+1}}\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {c}{n+1}}\int {\frac {x^{n+1}dx}{c^{2}+x^{2}}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$

${\displaystyle \int (ax+b)^{n}dx={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{( }}n\neq -1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{ax+b}}={\frac {1}{a}}\ln \left|ax+b\right|}$
${\displaystyle \int x(ax+b)^{n}dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{( }}n\not \in \{1,2\}{\mbox{)}}}$
${\displaystyle \int {\frac {x}{ax+b}}dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|}$
${\displaystyle \int {\frac {x}{(ax+b)^{2}}}dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|}$
${\displaystyle \int {\frac {x}{(ax+b)^{n}}}dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}\qquad {\mbox{( }}n\not \in \{1,2\}{\mbox{)}}}$
${\displaystyle \int {\frac {x^{2}}{ax+b}}dx={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)}$
${\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}dx={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)}$
${\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)}$
${\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx={\frac {1}{a^{3}}}\left(-{\frac {1}{(n-3)(ax+b)^{n-3}}}+{\frac {2b}{(n-2)(a+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right)\qquad {\mbox{( }}n\not \in \{1,2,3\}{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{x(ax+b)}}=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|}$
${\displaystyle \int {\frac {dx}{x^{2}(ax+b)}}=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|}$
${\displaystyle \int {\frac {dx}{x^{2}(ax+b)^{2}}}=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)}$
${\displaystyle \int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!}$
${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}\qquad {\mbox{( }}|x|<|a|{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{( }}|x|>|a|{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{( }}4ac-b^{2}>0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|\qquad {\mbox{( }}4ac-b^{2}<0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}=-{\frac {2}{2ax+b}}\qquad {\mbox{( }}4ac-b^{2}=0{\mbox{)}}}$
${\displaystyle \int {\frac {x}{ax^{2}+bx+c}}dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}}$
${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{( }}4ac-b^{2}>0{\mbox{)}}}$
${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{( }}4ac-b^{2}<0{\mbox{)}}}$
${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}\qquad {\mbox{( }}4ac-b^{2}=0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{(ax^{2}+bx+c)^{n}}}={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {dx}{(ax^{2}+bx+c)^{n-1}}}\,\!}$
${\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx={\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {dx}{(ax^{2}+bx+c)^{n-1}}}\,\!}$
${\displaystyle \int {\frac {dx}{x(ax^{2}+bx+c)}}={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {dx}{ax^{2}+bx+c}}}$

${\displaystyle \int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left(x+r\right)\right)}$
${\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {3}{8}}a^{2}xr+{\frac {3}{8}}a^{4}\ln \left(x+r\right)}$
${\displaystyle \int r^{5}\;dx={\frac {1}{6}}xr^{5}+{\frac {5}{24}}a^{2}xr^{3}+{\frac {5}{16}}a^{4}xr+{\frac {5}{16}}a^{6}\ln \left(x+r\right)}$
${\displaystyle \int xr\;dx={\frac {r^{3}}{3}}}$
${\displaystyle \int xr^{3}\;dx={\frac {r^{5}}{5}}}$
${\displaystyle \int xr^{2n+1}\;dx={\frac {r^{2n+3}}{2n+3}}}$
${\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left(x+r\right)}$
${\displaystyle \int x^{2}r^{3}\;dx={\frac {xr^{5}}{6}}-{\frac {a^{2}xr^{3}}{24}}-{\frac {a^{4}xr}{16}}-{\frac {a^{6}}{16}}\ln \left(x+r\right)}$
${\displaystyle \int x^{3}r\;dx={\frac {r^{5}}{5}}-{\frac {a^{2}r^{3}}{3}}}$
${\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{7}}{7}}-{\frac {a^{2}r^{5}}{5}}}$
${\displaystyle \int x^{3}r^{2n+1}\;dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}}$
${\displaystyle \int x^{4}r\;dx={\frac {x^{3}r^{3}}{6}}-{\frac {a^{2}xr^{3}}{8}}+{\frac {a^{4}xr}{16}}+{\frac {a^{6}}{16}}\ln \left(x+r\right)}$
${\displaystyle \int x^{4}r^{3}\;dx={\frac {x^{3}r^{5}}{8}}-{\frac {a^{2}xr^{5}}{16}}+{\frac {a^{4}xr^{3}}{64}}+{\frac {3a^{6}xr}{128}}+{\frac {3a^{8}}{128}}\ln \left(x+r\right)}$
${\displaystyle \int x^{5}r\;dx={\frac {r^{7}}{7}}-{\frac {2a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}}$
${\displaystyle \int x^{5}r^{3}\;dx={\frac {r^{9}}{9}}-{\frac {2a^{2}r^{7}}{7}}+{\frac {a^{4}r^{5}}{5}}}$
${\displaystyle \int x^{5}r^{2n+1}\;dx={\frac {r^{2n+7}}{2n+7}}-{\frac {2a^{2}r^{2n+5}}{2n+5}}+{\frac {a^{4}r^{2n+3}}{2n+3}}}$
${\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\,\operatorname {arsinh} {\frac {a}{x}}}$
${\displaystyle \int {\frac {r^{3}\;dx}{x}}={\frac {r^{3}}{3}}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|}$
${\displaystyle \int {\frac {r^{5}\;dx}{x}}={\frac {r^{5}}{5}}+{\frac {a^{2}r^{3}}{3}}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}}\right|}$
${\displaystyle \int {\frac {r^{7}\;dx}{x}}={\frac {r^{7}}{7}}+{\frac {a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}+a^{6}r-a^{7}\ln \left|{\frac {a+r}{x}}\right|}$
${\displaystyle \int {\frac {dx}{r}}=\operatorname {arsinh} {\frac {x}{a}}=\ln \left({\frac {x+r}{a}}\right)}$
${\displaystyle \int {\frac {dx}{r^{3}}}={\frac {x}{a^{2}r}}}$
${\displaystyle \int {\frac {x\,dx}{r}}=r}$
${\displaystyle \int {\frac {x\,dx}{r^{3}}}=-{\frac {1}{r}}}$
${\displaystyle \int {\frac {x^{2}\;dx}{r}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\,\operatorname {arsinh} {\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left({\frac {x+r}{a}}\right)}$
${\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\operatorname {arsinh} {\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|}$