7.1 Integration By Parts/28: Difference between revisions
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<math> \int_{1}^{2}\frac{(\ln{x})^2}{x^3} </math> | <math> \int_{1}^{2}\frac{(\ln{x})^2}{x^3} </math> | ||
<math> u = \ln{x} \qquad dv = \frac{1}{x^3} </math> <br><br> | <math> u = \ln^2{x} \qquad dv = \frac{1}{x^3} </math> <br><br> | ||
<math> du = \frac{ | <math> du = \frac{2\ln{(x)}}{x} \qquad v = -\frac{1}{2x^2} </math> | ||
<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] | ||
u = | |||
Revision as of 22:19, 16 December 2022
![{\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]u=&=-{\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cfe22d76ba73498adcbd612ad3b9af0169a99126)