តារាងអាំងតេក្រាលនៃអនុគមន៍សនិទាន

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

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

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

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

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

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

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

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

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

ចំពោះ $a\neq 0:$

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

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

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

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

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

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

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

គ្រប់អនុគមន៍សនិទានទាំងអស់អាច​ធ្វើអាំងតេក្រាលបានដោយប្រើសមីការនិងអាំងតេក្រាលដោយផ្នែក ដោយបំបែកអនុគមន៍សនិទានជា​ផលបូកនៃអនុគមន៍ដែលមានទំរង់៖

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