7.1 Integration By Parts/51b: Difference between revisions

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<math> \text{Use } \left[\int(\ln^{n})x(\ln{x})^n - n\int(\ln{x})^{n-1}dx\right] \text{ to evaluate } \int(\ln{x})^3dx</math> <br><br>
<math> \text{Use } \left[\int(\ln^{n}) = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx\right] \text{ to evaluate } \int(\ln{x})^3dx</math> <br><br>


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Revision as of 18:05, 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 } \left[\int(\ln^{n}) = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx\right] \text{ 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 \begin{align} \int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{ \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{align} }