ខាងក្រោមនេះជាតារាងអាំងតេក្រាល (ព្រីមីទីវ) នៃ អនុគមន៍លោការីត។ សំរាប់តារាងពេញលេញ សូមមើលនៅតារាងអាំងតេក្រាល។ ចំនាំ៖ នៅក្នុងអត្ថបទនេះ យើងសន្មតយក x > 0 {\displaystyle x>0\,} ។ ∫ ln c x d x = x ln c x − x {\displaystyle \int \ln cx\;dx=x\ln cx-x} ∫ ln ( a x + b ) d x = ( a x + b ) ln ( a x + b ) − ( a x + b ) a {\displaystyle \int \ln(ax+b)\;dx={\frac {(ax+b)\ln(ax+b)-(ax+b)}{a}}} ∫ ( ln x ) 2 d x = x ( ln x ) 2 − 2 x ln x + 2 x {\displaystyle \int (\ln x)^{2}\;dx=x(\ln x)^{2}-2x\ln x+2x} ∫ ( ln c x ) n d x = x ( ln c x ) n − n ∫ ( ln c x ) n − 1 d x {\displaystyle \int (\ln cx)^{n}\;dx=x(\ln cx)^{n}-n\int (\ln cx)^{n-1}dx} ∫ d x ln x = ln | ln x | + ln x + ∑ k = 2 ∞ ( ln x ) k k ⋅ k ! {\displaystyle \int {\frac {dx}{\ln x}}=\ln |\ln x|+\ln x+\sum _{k=2}^{\infty }{\frac {(\ln x)^{k}}{k\cdot k!}}} ∫ d x ( ln x ) n = − x ( n − 1 ) ( ln x ) n − 1 + 1 n − 1 ∫ d x ( ln x ) n − 1 {\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 } (ចំពោះ n ≠ 1 ) {\displaystyle n\neq 1)} ) ∫ x m ln x d x = x m + 1 ( ln x m + 1 − 1 ( m + 1 ) 2 ) {\displaystyle \int x^{m}\ln x\;dx=x^{m+1}\left({\frac {\ln x}{m+1}}-{\frac {1}{(m+1)^{2}}}\right)\qquad \ } (ចំពោះ m ≠ − 1 {\displaystyle m\neq -1\,} ) ∫ x m ( ln x ) n d x = x m + 1 ( ln x ) n m + 1 − n m + 1 ∫ x m ( ln x ) n − 1 d x {\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 } (ចំពោះ m ≠ − 1 {\displaystyle m\neq -1\,} ) ∫ ( ln x ) n d x x = ( ln x ) n + 1 n + 1 {\displaystyle \int {\frac {(\ln x)^{n}\;dx}{x}}={\frac {(\ln x)^{n+1}}{n+1}}\qquad } (ចំពោះ n ≠ − 1 {\displaystyle n\neq -1\,} ) ∫ ln x n d x x = ( ln x n ) 2 2 n {\displaystyle \int {\frac {\ln {x^{n}}\;dx}{x}}={\frac {(\ln {x^{n}})^{2}}{2n}}\qquad } (ចំពោះ n ≠ 0 {\displaystyle n\neq 0\,} ) ∫ ln x d x x m = − ln x ( m − 1 ) x m − 1 − 1 ( m − 1 ) 2 x m − 1 {\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 } (ចំពោះ m ≠ 1 {\displaystyle m\neq 1\,} ) ∫ ( ln x ) n d x x m = − ( ln x ) n ( m − 1 ) x m − 1 + n m − 1 ∫ ( ln x ) n − 1 d x x m {\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 } (ចំពោះ m ≠ 1 {\displaystyle m\neq 1\,} ) ∫ x m d x ( ln x ) n = − x m + 1 ( n − 1 ) ( ln x ) n − 1 + m + 1 n − 1 ∫ x m d x ( ln x ) n − 1 {\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 } (ចំពោះ n ≠ 1 {\displaystyle n\neq 1\,} ) ∫ d x x ln x = ln | ln x | {\displaystyle \int {\frac {dx}{x\ln x}}=\ln \left|\ln x\right|} ∫ d x x n ln x = ln | ln x | + ∑ k = 1 ∞ ( − 1 ) k ( n − 1 ) k ( ln x ) k k ⋅ k ! {\displaystyle \int {\frac {dx}{x^{n}\ln x}}=\ln \left|\ln x\right|+\sum _{k=1}^{\infty }(-1)^{k}{\frac {(n-1)^{k}(\ln x)^{k}}{k\cdot k!}}} ∫ d x x ( ln x ) n = − 1 ( n − 1 ) ( ln x ) n − 1 {\displaystyle \int {\frac {dx}{x(\ln x)^{n}}}=-{\frac {1}{(n-1)(\ln x)^{n-1}}}\qquad } (ចំពោះ n ≠ 1 {\displaystyle n\neq 1\,} ) ∫ ln ( x 2 + a 2 ) d x = x ln ( x 2 + a 2 ) − 2 x + 2 a tan − 1 x a {\displaystyle \int \ln(x^{2}+a^{2})\;dx=x\ln(x^{2}+a^{2})-2x+2a\tan ^{-1}{\frac {x}{a}}} ∫ x x 2 + a 2 ln ( x 2 + a 2 ) d x = 1 4 ln 2 ( x 2 + a 2 ) {\displaystyle \int {\frac {x}{x^{2}+a^{2}}}\ln(x^{2}+a^{2})\;dx={\frac {1}{4}}\ln ^{2}(x^{2}+a^{2})} ∫ sin ( ln x ) d x = x 2 ( sin ( ln x ) − cos ( ln x ) ) {\displaystyle \int \sin(\ln x)\;dx={\frac {x}{2}}(\sin(\ln x)-\cos(\ln x))} ∫ cos ( ln x ) d x = x 2 ( sin ( ln x ) + cos ( ln x ) ) {\displaystyle \int \cos(\ln x)\;dx={\frac {x}{2}}(\sin(\ln x)+\cos(\ln x))} ∫ e x ( x ln x − x − 1 x ) d x = e x ( x ln x − x − ln x ) {\displaystyle \int e^{x}\left(x\ln x-x-{\frac {1}{x}}\right)\;dx=e^{x}(x\ln x-x-\ln x)} ∫ 1 e x ( 1 x − ln x ) d x = ln x e x {\displaystyle \int {\frac {1}{e^{x}}}\left({\frac {1}{x}}-\ln x\right)\;dx={\frac {\ln x}{e^{x}}}} ∫ e x ( 1 ln x − 1 x ln 2 x ) d x = e x ln x {\displaystyle \int e^{x}\left({\frac {1}{\ln x}}-{\frac {1}{x\ln ^{2}x}}\right)\;dx={\frac {e^{x}}{\ln x}}} Remove adsLoading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads