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}\\[2ex]


& = -\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 22:17, 16 December 2022



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 ^{2}{(x)}}{2x^{2}}}-\int -{\frac {\ln {(x)}}{x^{3}}}&&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}+\int {\frac {\ln {(x)}}{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{aligned}}}