7.1 Integration By Parts/51b: Difference between revisions

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} \\ [1ex]
} \\ [1ex]


&= x\ln(x)^3 -3\left[\ln^{2}{(x)}\cdot x - 2\int\ln{(x)}dx\right] \\ [1ex]
&= x\ln^{3}(x) -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}_{
&= x\ln^{3}(x) -3x\ln^{2}{(x)} + \underbrace{6\int\ln{(x)}dx}_{
\begin{aligned}
\begin{aligned}
u & = \ln{(x)}  & dv &= dx \\[0.6ex]
u & = \ln{(x)}  & dv &= dx \\[0.6ex]
Line 21: Line 21:
\end{aligned}} \\ [1ex]
\end{aligned}} \\ [1ex]


&= x\ln(x)^3 -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int dx\right] \\[1ex]
&= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int dx\right] \\[1ex]
&= x\ln(x)^3 -3x\ln^{2}{(x)} + 6x\ln{(x)} - 6x + C
&= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6x\ln{(x)} - 6x + C





Revision as of 18:01, 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 ^{3}(x)-3\left[\ln ^{2}{(x)}\cdot x-2\int \ln {(x)}dx\right]\\[1ex]&=x\ln ^{3}(x)-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 ^{3}(x)-3x\ln ^{2}{(x)}+6\left[\ln {(x)}\cdot x-\int dx\right]\\[1ex]&=x\ln ^{3}(x)-3x\ln ^{2}{(x)}+6x\ln {(x)}-6x+C\end{aligned}}}