តារាងអាំងតេក្រាលនៃអនុគមន៍សនិទាន

ខាងក្រោមនេះជា​តារាងអាំងតេក្រាល (ព្រីមីទីវ) នៃអនុគមន៍សនិទាន។ សូមមើល តារាងអាំងតេក្រាល សំរាប់បញ្ជីពេញលេញនៃគ្រប់អាំងតេក្រាល។

${\displaystyle \int (ax+b)^{n}dx}$	${\displaystyle ={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad }$   (ចំពោះ${\displaystyle n\neq -1\,\!}$)
${\displaystyle \int {\frac {c}{ax+b}}dx}$	${\displaystyle ={\frac {c}{a}}\ln \left|ax+b\right|}$
${\displaystyle \int x(ax+b)^{n}dx}$	${\displaystyle ={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad }$   (ចំពោះ${\displaystyle n\not \in \{-1,-2\}\,}$)

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

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

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

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

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

${\displaystyle \int {\frac {1}{x^{2}+a^{2}}}dx={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!}$

${\displaystyle \int {\frac {1}{x^{2}-a^{2}}}dx={\begin{cases}-{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}&(|x|<|a|)\\-{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}&(|x|>|a|)\end{cases}}}$

ចំពោះ ${\displaystyle a\neq 0:}$

${\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}{\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}&(4ac-b^{2}>0)\\-{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|&(4ac-b^{2}<0)\\-{\frac {2}{2ax+b}}\qquad (4ac-b^{2}=0)\end{cases}}}$

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

${\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\begin{cases}{\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}&(4ac-b^{2}>0)\\{\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}&(4ac-b^{2}<0)\\{\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}&(4ac-b^{2}=0)\end{cases}}}$

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

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

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

${\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin({\frac {(2k-1)\pi }{2^{n}}})\arctan[\left(x-\cos({\frac {(2k-1)\pi }{2^{n}}})\right)\csc({\frac {(2k-1)\pi }{2^{n}}})]\right]-{\frac {1}{2^{n}}}\left[\cos({\frac {(2k-1)\pi }{2^{n}}})\ln \left|x^{2}-2x\cos({\frac {(2k-1)\pi }{2^{n}}})+1\right|\right]\right\}}$

គ្រប់អនុគមន៍សនិទានទាំងអស់អាច​ធ្វើអាំងតេក្រាលបានដោយប្រើសមីការនិងអាំងតេក្រាលដោយផ្នែក ដោយបំបែកអនុគមន៍សនិទានជា​ផលបូកនៃអនុគមន៍ដែលមានទំរង់៖

${\displaystyle {\frac {ex+f}{\left(ax^{2}+bx+c\right)^{n}}}}$