7.1 Integration By Parts/51b: Difference between revisions

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\end{aligned}} \\ [1ex]
\end{aligned}} \\ [1ex]


&= x\ln(x)^3 -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int\dx\right]
&= x\ln(x)^3 -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int dx\right]


\end{align}
\end{align}


</math>
</math>

Revision as of 17:59, 29 November 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 \text{Use exercise 47 to evaluate} \int(\ln{x})^3dx }

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 \text{Exercise 47: } x(\ln{x})^n-n\int(\ln{x})^{n-1}dx }

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int \ln(x)^{3}dx&=x\ln(x)^{3}-\underbrace {3\int \ln(x)^{2}dx} _{\begin{aligned}u&=\ln ^{2}{(x)}&dv&=dx\\[0.6ex]du&={\tfrac {2\ln {(x)}}{x}}dx&v&=x\end{aligned}}\\[1ex]&=x\ln(x)^{3}-3\left[\ln ^{2}{(x)}\cdot x-2\int \ln {(x)}dx\right]\\[1ex]&=x\ln(x)^{3}-3x\ln ^{2}{(x)}+\underbrace {6\int \ln {(x)}dx} _{\begin{aligned}u&=\ln {(x)}&dv&=dx\\[0.6ex]du&={\tfrac {1}{x}}dx&v&=x\end{aligned}}\\[1ex]&=x\ln(x)^{3}-3x\ln ^{2}{(x)}+6\left[\ln {(x)}\cdot x-\int dx\right]\end{aligned}}}