Tích phân hàm số toán lủy thừa của cosecant

${\displaystyle \int \csc {ax}\,dx=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C}$
${\displaystyle \int \csc ^{2}{x}\,dx=-\cot {x}+C}$
${\displaystyle \int \csc ^{n}{ax}\,dx=-{\frac {\csc ^{n-1}{ax}\csc {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{ax}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,}$
${\displaystyle \int {\frac {dx}{\csc {x}+1}}=x-{\frac {2\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}+\sin {\frac {x}{2}}}}+C}$
${\displaystyle \int {\frac {dx}{\csc {x}-1}}={\frac {2\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}-\sin {\frac {x}{2}}}}-x+C}$

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

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

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

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

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

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

${\displaystyle \int \sinh cx\,dx={\frac {1}{c}}\cosh cx}$
${\displaystyle \int \cosh cx\,dx={\frac {1}{c}}\sinh cx}$
${\displaystyle \int \sinh ^{2}cx\,dx={\frac {1}{4c}}\sinh 2cx-{\frac {x}{2}}}$
${\displaystyle \int \cosh ^{2}cx\,dx={\frac {1}{4c}}\sinh 2cx+{\frac {x}{2}}}$
${\displaystyle \int \sinh ^{n}cx\,dx={\frac {1}{cn}}\sinh ^{n-1}cx\cosh cx-{\frac {n-1}{n}}\int \sinh ^{n-2}cx\,dx\qquad {\mbox{(}}n>0{\mbox{)}}}$
hay: ${\displaystyle \int \sinh ^{n}cx\,dx={\frac {1}{c(n+1)}}\sinh ^{n+1}cx\cosh cx-{\frac {n+2}{n+1}}\int \sinh ^{n+2}cx\,dx\qquad {\mbox{(}}n<0{\mbox{, }}n\neq -1{\mbox{)}}}$
${\displaystyle \int \cosh ^{n}cx\,dx={\frac {1}{cn}}\sinh cx\cosh ^{n-1}cx+{\frac {n-1}{n}}\int \cosh ^{n-2}cx\,dx\qquad {\mbox{(}}n>0{\mbox{)}}}$
hay: ${\displaystyle \int \cosh ^{n}cx\,dx=-{\frac {1}{c(n+1)}}\sinh cx\cosh ^{n+1}cx-{\frac {n+2}{n+1}}\int \cosh ^{n+2}cx\,dx\qquad {\mbox{(}}n<0{\mbox{, }}n\neq -1{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|\tanh {\frac {cx}{2}}\right|}$
hay: ${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|{\frac {\cosh cx-1}{\sinh cx}}\right|}$
hay: ${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|{\frac {\sinh cx}{\cosh cx+1}}\right|}$
hay: ${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|{\frac {\cosh cx-1}{\cosh cx+1}}\right|}$
${\displaystyle \int {\frac {dx}{\cosh cx}}={\frac {2}{c}}\arctan e^{cx}}$
${\displaystyle \int {\frac {dx}{\sinh ^{n}cx}}={\frac {\cosh cx}{c(n-1)\sinh ^{n-1}cx}}-{\frac {n-2}{n-1}}\int {\frac {dx}{\sinh ^{n-2}cx}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{\cosh ^{n}cx}}={\frac {\sinh cx}{c(n-1)\cosh ^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cosh ^{n-2}cx}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {\cosh ^{n}cx}{\sinh ^{m}cx}}dx={\frac {\cosh ^{n-1}cx}{c(n-m)\sinh ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {\cosh ^{n-2}cx}{\sinh ^{m}cx}}dx\qquad {\mbox{(}}m\neq n{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\cosh ^{n}cx}{\sinh ^{m}cx}}dx=-{\frac {\cosh ^{n+1}cx}{c(m-1)\sinh ^{m-1}cx}}+{\frac {n-m+2}{m-1}}\int {\frac {\cosh ^{n}cx}{\sinh ^{m-2}cx}}dx\qquad {\mbox{(}}m\neq 1{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\cosh ^{n}cx}{\sinh ^{m}cx}}dx=-{\frac {\cosh ^{n-1}cx}{c(m-1)\sinh ^{m-1}cx}}+{\frac {n-1}{m-1}}\int {\frac {\cosh ^{n-2}cx}{\sinh ^{m-2}cx}}dx\qquad {\mbox{(}}m\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {\sinh ^{m}cx}{\cosh ^{n}cx}}dx={\frac {\sinh ^{m-1}cx}{c(m-n)\cosh ^{n-1}cx}}+{\frac {m-1}{m-n}}\int {\frac {\sinh ^{m-2}cx}{\cosh ^{n}cx}}dx\qquad {\mbox{(}}m\neq n{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\sinh ^{m}cx}{\cosh ^{n}cx}}dx={\frac {\sinh ^{m+1}cx}{c(n-1)\cosh ^{n-1}cx}}+{\frac {m-n+2}{n-1}}\int {\frac {\sinh ^{m}cx}{\cosh ^{n-2}cx}}dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\sinh ^{m}cx}{\cosh ^{n}cx}}dx=-{\frac {\sinh ^{m-1}cx}{c(n-1)\cosh ^{n-1}cx}}+{\frac {m-1}{n-1}}\int {\frac {\sinh ^{m-2}cx}{\cosh ^{n-2}cx}}dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int x\sinh cx\,dx={\frac {1}{c}}x\cosh cx-{\frac {1}{c^{2}}}\sinh cx}$
${\displaystyle \int x\cosh cx\,dx={\frac {1}{c}}x\sinh cx-{\frac {1}{c^{2}}}\cosh cx}$
${\displaystyle \int \tanh cx\,dx={\frac {1}{c}}\ln |\cosh cx|}$
${\displaystyle \int \coth cx\,dx={\frac {1}{c}}\ln |\sinh cx|}$
${\displaystyle \int \tanh ^{n}cx\,dx=-{\frac {1}{c(n-1)}}\tanh ^{n-1}cx+\int \tanh ^{n-2}cx\,dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int \coth ^{n}cx\,dx=-{\frac {1}{c(n-1)}}\coth ^{n-1}cx+\int \coth ^{n-2}cx\,dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int \sinh bx\sinh cx\,dx={\frac {1}{b^{2}-c^{2}}}(b\sinh cx\cosh bx-c\cosh cx\sinh bx)\qquad {\mbox{(}}b^{2}\neq c^{2}{\mbox{)}}}$
${\displaystyle \int \cosh bx\cosh cx\,dx={\frac {1}{b^{2}-c^{2}}}(b\sinh bx\cosh cx-c\sinh cx\cosh bx)\qquad {\mbox{(}}b^{2}\neq c^{2}{\mbox{)}}}$
${\displaystyle \int \cosh bx\sinh cx\,dx={\frac {1}{b^{2}-c^{2}}}(b\sinh bx\sinh cx-c\cosh bx\cosh cx)\qquad {\mbox{(}}b^{2}\neq c^{2}{\mbox{)}}}$
${\displaystyle \int \sinh(ax+b)\sin(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\cosh(ax+b)\sin(cx+d)-{\frac {c}{a^{2}+c^{2}}}\sinh(ax+b)\cos(cx+d)}$
${\displaystyle \int \sinh(ax+b)\cos(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\cosh(ax+b)\cos(cx+d)+{\frac {c}{a^{2}+c^{2}}}\sinh(ax+b)\sin(cx+d)}$
${\displaystyle \int \cosh(ax+b)\sin(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\sinh(ax+b)\sin(cx+d)-{\frac {c}{a^{2}+c^{2}}}\cosh(ax+b)\cos(cx+d)}$
${\displaystyle \int \cosh(ax+b)\cos(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\sinh(ax+b)\cos(cx+d)+{\frac {c}{a^{2}+c^{2}}}\cosh(ax+b)\sin(cx+d)}$

${\displaystyle \int \mathrm {arsinh} \,{\frac {x}{c}}\,dx=x\,\mathrm {arsinh} \,{\frac {x}{c}}-{\sqrt {x^{2}+c^{2}}}}$
${\displaystyle \int \mathrm {arcosh} \,{\frac {x}{c}}\,dx=x\,\mathrm {arcosh} \,{\frac {x}{c}}-{\sqrt {x^{2}-c^{2}}}}$
${\displaystyle \int \mathrm {artanh} \,{\frac {x}{c}}\,dx=x\,\mathrm {artanh} \,{\frac {x}{c}}+{\frac {c}{2}}\ln |c^{2}-x^{2}|\qquad {\mbox{(}}|x|<|c|{\mbox{)}}}$
${\displaystyle \int \mathrm {arcoth} \,{\frac {x}{c}}\,dx=x\,\mathrm {arcoth} \,{\frac {x}{c}}+{\frac {c}{2}}\ln |x^{2}-c^{2}|\qquad {\mbox{(}}|x|>|c|{\mbox{)}}}$
${\displaystyle \int \mathrm {arsech} \,{\frac {x}{c}}\,dx=x\,\mathrm {arsech} \,{\frac {x}{c}}-c\,\mathrm {arctan} \,{\frac {x\,{\sqrt {\frac {c-x}{c+x}}}}{x-c}}\qquad {\mbox{(}}x\in (0,\,c){\mbox{)}}}$
${\displaystyle \int \mathrm {arcsch} \,{\frac {x}{c}}\,dx=x\,\mathrm {arcsch} \,{\frac {x}{c}}+c\,\ln \,{\frac {x+{\sqrt {x^{2}+c^{2}}}}{c}}\qquad {\mbox{(}}x\in (0,\,c){\mbox{)}}}$
${\displaystyle \int e^{cx}\;dx={\frac {1}{c}}e^{cx}}$
${\displaystyle \int a^{cx}\;dx={\frac {1}{c\ln a}}a^{cx}\qquad {\mbox{(}}a>0,{\mbox{ }}a\neq 1{\mbox{)}}}$
${\displaystyle \int xe^{cx}\;dx={\frac {e^{cx}}{c^{2}}}(cx-1)}$
${\displaystyle \int x^{2}e^{cx}\;dx=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)}$
${\displaystyle \int x^{n}e^{cx}\;dx={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}dx}$
${\displaystyle \int {\frac {e^{cx}\;dx}{x}}=\ln |x|+\sum _{i=1}^{\infty }{\frac {(cx)^{i}}{i\cdot i!}}}$
${\displaystyle \int {\frac {e^{cx}\;dx}{x^{n}}}={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,dx\right)\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int e^{cx}\ln x\;dx={\frac {1}{c}}e^{cx}\ln |x|-\operatorname {Ei} \,(cx)}$
${\displaystyle \int e^{cx}\sin bx\;dx={\frac {e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)}$
${\displaystyle \int e^{cx}\cos bx\;dx={\frac {e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)}$
${\displaystyle \int e^{cx}\sin ^{n}x\;dx={\frac {e^{cx}\sin ^{n-1}x}{c^{2}+n^{2}}}(c\sin x-n\cos x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\sin ^{n-2}x\;dx}$
${\displaystyle \int e^{cx}\cos ^{n}x\;dx={\frac {e^{cx}\cos ^{n-1}x}{c^{2}+n^{2}}}(c\cos x+n\sin x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\cos ^{n-2}x\;dx}$
${\displaystyle \int xe^{cx^{2}}\;dx={\frac {1}{2c}}\;e^{cx^{2}}}$
${\displaystyle \int {1 \over \sigma {\sqrt {2\pi }}}\,e^{-{(x-\mu )^{2}/2\sigma ^{2}}}\;dx={\frac {1}{2\sigma }}(1+{\mbox{erf}}\,{\frac {x-\mu }{\sigma {\sqrt {2}}}})}$
${\displaystyle \int e^{x^{2}}\,dx=e^{x^{2}}\left(\sum _{j=0}^{n-1}c_{2j}\,{\frac {1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int {\frac {e^{x^{2}}}{x^{2n}}}\;dx\quad (n>0),}$
với ${\displaystyle c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)\,!}{j!\,2^{2j+1}}}\ .}$
${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\sqrt {\pi \over a}}}$
${\displaystyle \int _{0}^{\infty }x^{2n}e^{-{x^{2}}/{a^{2}}}\,dx={\sqrt {\pi }}{(2n)! \over {n!}}{\left({\frac {a}{2}}\right)}^{2n+1}}$
${\displaystyle \int \ln cx\,dx=x\ln cx-x}$
${\displaystyle \int (\ln x)^{2}\;dx=x(\ln x)^{2}-2x\ln x+2x}$
${\displaystyle \int (\ln cx)^{n}\;dx=x(\ln cx)^{n}-n\int (\ln cx)^{n-1}dx}$
${\displaystyle \int {\frac {dx}{\ln x}}=\ln |\ln x|+\ln x+\sum _{i=2}^{\infty }{\frac {(\ln x)^{i}}{i\cdot i!}}}$
${\displaystyle \int {\frac {dx}{(\ln x)^{n}}}=-{\frac {x}{(n-1)(\ln x)^{n-1}}}+{\frac {1}{n-1}}\int {\frac {dx}{(\ln x)^{n-1}}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int x^{m}\ln x\;dx=x^{m+1}\left({\frac {\ln x}{m+1}}-{\frac {1}{(m+1)^{2}}}\right)\qquad {\mbox{(}}m\neq -1{\mbox{)}}}$
${\displaystyle \int x^{m}(\ln x)^{n}\;dx={\frac {x^{m+1}(\ln x)^{n}}{m+1}}-{\frac {n}{m+1}}\int x^{m}(\ln x)^{n-1}dx\qquad {\mbox{(}}m\neq -1{\mbox{)}}}$
${\displaystyle \int {\frac {(\ln x)^{n}\;dx}{x}}={\frac {(\ln x)^{n+1}}{n+1}}\qquad {\mbox{(}}n\neq -1{\mbox{)}}}$
${\displaystyle \int {\frac {\ln x\,dx}{x^{m}}}=-{\frac {\ln x}{(m-1)x^{m-1}}}-{\frac {1}{(m-1)^{2}x^{m-1}}}\qquad {\mbox{(}}m\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {(\ln x)^{n}\;dx}{x^{m}}}=-{\frac {(\ln x)^{n}}{(m-1)x^{m-1}}}+{\frac {n}{m-1}}\int {\frac {(\ln x)^{n-1}dx}{x^{m}}}\qquad {\mbox{( }}m\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {x^{m}\;dx}{(\ln x)^{n}}}=-{\frac {x^{m+1}}{(n-1)(\ln x)^{n-1}}}+{\frac {m+1}{n-1}}\int {\frac {x^{m}dx}{(\ln x)^{n-1}}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{x\ln x}}=\ln |\ln x|}$
${\displaystyle \int {\frac {dx}{x^{n}\ln x}}=\ln |\ln x|+\sum _{i=1}^{\infty }(-1)^{i}{\frac {(n-1)^{i}(\ln x)^{i}}{i\cdot i!}}}$
${\displaystyle \int {\frac {dx}{x(\ln x)^{n}}}=-{\frac {1}{(n-1)(\ln x)^{n-1}}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int \sin(\ln x)\;dx={\frac {x}{2}}(\sin(\ln x)-\cos(\ln x))}$
${\displaystyle \int \cos(\ln x)\;dx={\frac {x}{2}}(\sin(\ln x)+\cos(\ln x))}$

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