តារាងអាំងតេក្រាលនៃអនុគមន៍លោការីត

ចំនាំ៖ នៅក្នុងអត្ថបទនេះ យើងសន្មតយក $x > 0 \,$ ។

$\int\ln cx\;dx = x\ln cx - x$

$\int\ln (ax + b)\;dx = \frac{(ax+b)\ln(ax+b) - (ax+b)}{a}$

$\int (\ln x)^2\; dx = x(\ln x)^2 - 2x\ln x + 2x$

$\int (\ln cx)^n\; dx = x(\ln cx)^n - n\int (\ln cx)^{n-1} dx$

$\int \frac{dx}{\ln x} = \ln|\ln x| + \ln x + \sum^\infty_{k=2}\frac{(\ln x)^k}{k\cdot k!}$

$\int \frac{dx}{(\ln x)^n} = -\frac{x}{(n-1)(\ln x)^{n-1}} + \frac{1}{n-1}\int\frac{dx}{(\ln x)^{n-1}} \qquad$ (ចំពោះ $n\neq 1)$ )

$\int x^m\ln x\;dx = x^{m+1}\left(\frac{\ln x}{m+1}-\frac{1}{(m+1)^2}\right) \qquad\$ (ចំពោះ $m\neq -1\,$ )

$\int x^m (\ln x)^n\; dx = \frac{x^{m+1}(\ln x)^n}{m+1} - \frac{n}{m+1}\int x^m (\ln x)^{n-1} dx \qquad$ (ចំពោះ $m\neq -1\,$ )

$\int \frac{(\ln x)^n\; dx}{x} = \frac{(\ln x)^{n+1}}{n+1} \qquad$ (ចំពោះ $n\neq -1\,$ )

$\int \frac{\ln{x^n}\;dx}{x} = \frac{(\ln{x^n})^2}{2n} \qquad$ (ចំពោះ $n\neq 0 \,$ )

$\int \frac{\ln x\,dx}{x^m} = -\frac{\ln x}{(m-1)x^{m-1}}-\frac{1}{(m-1)^2 x^{m-1}} \qquad$ (ចំពោះ $m\neq 1\,$ )

$\int \frac{(\ln x)^n\; dx}{x^m} = -\frac{(\ln x)^n}{(m-1)x^{m-1}} + \frac{n}{m-1}\int\frac{(\ln x)^{n-1} dx}{x^m} \qquad$ (ចំពោះ $m\neq 1\,$ )

$\int \frac{x^m\; dx}{(\ln x)^n} = -\frac{x^{m+1}}{(n-1)(\ln x)^{n-1}} + \frac{m+1}{n-1}\int\frac{x^m dx}{(\ln x)^{n-1}} \qquad$ (ចំពោះ $n\neq 1\,$ )

$\int \frac{dx}{x\ln x} = \ln \left|\ln x\right|$

$\int \frac{dx}{x^n\ln x} = \ln \left|\ln x\right| + \sum^\infty_{k=1} (-1)^k\frac{(n-1)^k(\ln x)^k}{k\cdot k!}$

$\int \frac{dx}{x (\ln x)^n} = -\frac{1}{(n-1)(\ln x)^{n-1}} \qquad$ (ចំពោះ $n\neq 1\,$ )

$\int \ln(x^2+a^2)\; dx = x\ln(x^2+a^2)-2x+2a\tan^{-1} \frac{x}{a}$

$\int \frac{x}{x^2+a^2}\ln(x^2+a^2)\; dx = \frac{1}{4} \ln^2(x^2+a^2)$

$\int \sin (\ln x)\;dx = \frac{x}{2}(\sin (\ln x) - \cos (\ln x))$

$\int \cos (\ln x)\;dx = \frac{x}{2}(\sin (\ln x) + \cos (\ln x))$

$\int e^x \left(x \ln x - x - \frac{1}{x}\right)\;dx = e^x (x \ln x - x - \ln x)$

$\int \frac{1}{e^x} \left( \frac{1}{x}-\ln x \right)\;dx = \frac{\ln x}{e^x}$

$\int e^x \left( \frac{1}{\ln x}- \frac{1}{x\ln^2 x} \right)\;dx = \frac{e^x}{\ln x}$