7.1 Integration By Parts/28: Difference between revisions
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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} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 22:05, 16 December 2022
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 \int_{1}^{2}\frac{(\ln{x})^2}{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 u = \ln{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{1}{x} \qquad v = -\frac{1}{2x^2} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {\ln {x}^{2}}{x^{3}}}&=-{\frac {\ln {(x)}}{2x^{2}}}-\int -{\frac {1}{2x^{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}}}\end{aligned}}}