فرمول های انتگرال (Integral) شامل ln (لگاریتم طبیعی - Natural Logarithm)، در ریاضیات (Mathematics)
\[ \int \ln ax\ dx = x \ln ax - x \] \[ \int x \ln x \ dx = \frac{1}{2} x^2 \ln x-\frac{x^2}{4} \] \[ \int x^2 \ln x \ dx = \frac{1}{3} x^3 \ln x-\frac{x^3}{9} \] \[ \int x^n \ln x\ dx = x^{n+1}\left( \dfrac{\ln x}{n+1}-\dfrac{1}{(n+1)^2}\right),\hspace{2ex} n\neq -1 \] \[ \int {{x^n}} {(\ln ax)^m}dx = {{{x^{n + 1}}{{(\ln ax)}^m}} \over {n + 1}} - {m \over {n + 1}}\int {{x^n}} {(\ln ax)^{m - 1}}dx \ \ \ \ ,n \ne - 1 \] \[ \int {{x^{ - 1}}} {(\ln ax)^m}dx = {{{{(\ln ax)}^{m + 1}}} \over {m + 1}} + C \ \ \ \ ,m \ne - 1 \] \[ \int {{{dx} \over {x\ln ax}}} = \ln |\ln ax| + C \] \[ \int \frac{\ln ax}{x}\ dx = \frac{1}{2}\left ( \ln ax \right)^2 \] \[ \int \frac{\ln x}{x^2}\ dx = -\frac{1}{x}-\frac{\ln x}{x} \] \[ \int \ln (ax + b) \ dx = \left ( x + \frac{b}{a} \right) \ln (ax+b) - x , a\ne 0 \] \[ \int \ln ( x^2 + a^2 )\hspace{.5ex} {dx} = x \ln (x^2 + a^2 ) +2a\tan^{-1} \frac{x}{a} - 2x \] \[ \int \ln ( x^2 - a^2 )\hspace{.5ex} {dx} = x \ln (x^2 - a^2 ) +a\ln \frac{x+a}{x-a} - 2x \] \[ \int \ln \left ( ax^2 + bx + c\right) \ dx = \frac{1}{a}\sqrt{4ac-b^2}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}} -2x + \left( \frac{b}{2a}+x \right )\ln \left (ax^2+bx+c \right) \] \[ \int x \ln (ax + b)\ dx = \frac{bx}{2a}-\frac{1}{4}x^2 +\frac{1}{2}\left(x^2-\frac{b^2}{a^2}\right)\ln (ax+b) \] \[ \int x \ln \left ( a^2 - b^2 x^2 \right )\ dx = -\frac{1}{2}x^2+ \frac{1}{2}\left( x^2 - \frac{a^2}{b^2} \right ) \ln \left (a^2 -b^2 x^2 \right) \] \[ \int (\ln x)^2\ dx = 2x - 2x \ln x + x (\ln x)^2 \] \[ \int (\ln x)^3\ dx = -6 x+x (\ln x)^3-3 x (\ln x)^2+6 x \ln x \] \[ \int x (\ln x)^2\ dx = \frac{x^2}{4}+\frac{1}{2} x^2 (\ln x)^2-\frac{1}{2} x^2 \ln x \] \[ \int x^2 (\ln x)^2\ dx = \frac{2 x^3}{27}+\frac{1}{3} x^3 (\ln x)^2-\frac{2}{9} x^3 \ln x \]
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Thomas' Calculus Early Transcendentals - George B. Thomas Jr., Maurice D. Weir, Joel R. Hass - 13th Edition
نویسنده
علیرضا گلمکانی
شماره کلید
10087
گزینه ها
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