7.1 Integration By Parts/28: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int_{1}^{2}\frac{\ln{x}^2}{x^3} = -\frac{\ln^2{(x)}}{2x^2} - \int-\frac{\ln{(x)}}{x^3} & = -\frac{\ln^2{(x)}}{2x^2} + \int\frac{\ln{(x)}}{x^3} \\[2ex] | \int_{1}^{2}\frac{\ln{x}^2}{x^3} = -\frac{\ln^2{(x)}}{2x^2} - \int-\frac{\ln{(x)}}{x^3} & = -\frac{\ln^2{(x)}}{2x^2} + \int\frac{\ln{(x)}}{x^3} && = -\frac{\ln{(x)}}{2x^2} - \int-frac{1}{2x^3} \\[2ex] | ||
& u = \ln{(x)} \qquad dv = \frac{1}{x^3} \\[2ex] | & u = \ln{(x)} \qquad dv = \frac{1}{x^3} \\[2ex] | ||
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& = -\frac{\ln{(x)}}{2x^2} - \frac{1}{2}\int\frac{1}{x^3} \\[2ex] | && = -\frac{\ln{(x)}}{2x^2} - \frac{1}{2}\int\frac{1}{x^3} \\[2ex] | ||
& = -\frac{\ln{(x)}}{2x^2} - \frac{1}{4x^2} | && = -\frac{\ln{(x)}}{2x^2} - \frac{1}{4x^2} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 22:26, 16 December 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}{\frac {(\ln {x})^{2}}{x^{3}}}}
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\ln ^{2}{x}\qquad dv={\frac {1}{x^{3}}}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle du = \frac{2\ln{(x)}}{x} \qquad v = -\frac{1}{2x^2} }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{1}^{2}\frac{\ln{x}^2}{x^3} = -\frac{\ln^2{(x)}}{2x^2} - \int-\frac{\ln{(x)}}{x^3} & = -\frac{\ln^2{(x)}}{2x^2} + \int\frac{\ln{(x)}}{x^3} && = -\frac{\ln{(x)}}{2x^2} - \int-frac{1}{2x^3} \\[2ex] & u = \ln{(x)} \qquad dv = \frac{1}{x^3} \\[2ex] & du = \frac{1}{x} \qquad v = -\frac{1}{2x^2} \\[2ex] && = -\frac{\ln{(x)}}{2x^2} - \frac{1}{2}\int\frac{1}{x^3} \\[2ex] && = -\frac{\ln{(x)}}{2x^2} - \frac{1}{4x^2} \end{align} }