7.1 Integration By Parts/51b - Revision history

Dvaezazizi@laalliance.org at 18:21, 29 November 2022

2022-11-29T18:21:22Z

← Older revision		Revision as of 18:21, 29 November 2022
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	\begin{align}		\begin{align}

	\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace~~[10][20]~~{3\int\ln(x)^2dx}_{		\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{
	\begin{aligned}		\begin{aligned}
	u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]		u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]

2022-11-29T18:21:12Z

← Older revision		Revision as of 18:21, 29 November 2022
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	\begin{align}		\begin{align}

	\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace[3]{3\int\ln(x)^2dx}_{		\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace[10][20]{3\int\ln(x)^2dx}_{
	\begin{aligned}		\begin{aligned}
	u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]		u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]

2022-11-29T18:20:51Z

← Older revision		Revision as of 18:20, 29 November 2022
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	\begin{align}		\begin{align}

	\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{		\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace[3]{3\int\ln(x)^2dx}_{
	\begin{aligned}		\begin{aligned}
	u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]		u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]

2022-11-29T18:16:06Z

← Older revision		Revision as of 18:16, 29 November 2022
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	&= 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)} + 6\left[\ln{(x)}\cdot x - \int dx\right] \\[1ex]
	&= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6x\ln{(x)} - 6x + C		&= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6x\ln{(x)} - 6x + C


	\end{align}		\end{align}

	</math>		</math>

2022-11-29T18:15:34Z

← Older revision		Revision as of 18:15, 29 November 2022
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	\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{		\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{

	\begin{aligned}		\begin{aligned}
	u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]		u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex]
	du & = \tfrac{2\ln{(x)}}{x}dx & v &= x		du & = \tfrac{2\ln{(x)}}{x}dx & v &= x
	\end{aligned}		\end{aligned}} \\ [2ex]
	} \\ [2ex]

	&= x\ln^{3}(x) -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^{3}(x) -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}

2022-11-29T18:14:07Z

← Older revision		Revision as of 18:14, 29 November 2022
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	du & = \tfrac{2\ln{(x)}}{x}dx & v &= x		du & = \tfrac{2\ln{(x)}}{x}dx & v &= x
	\end{aligned}		\end{aligned}
	} \\ [~~1ex~~]		} \\ [2ex]

	&= x\ln^{3}(x) -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]
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	u & = \ln{(x)} & dv &= dx \\[0.6ex]		u & = \ln{(x)} & dv &= dx \\[0.6ex]
	du & = \tfrac{1}{x}dx & v &= x		du & = \tfrac{1}{x}dx & v &= x
	\end{aligned}} \\ [~~1ex~~]		\end{aligned}} \\ [2ex]

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

2022-11-29T18:10:17Z

Protected "7.1 Integration By Parts/51b" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 18:10, 29 November 2022
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2022-11-29T18:08:07Z

← Older revision		Revision as of 18:08, 29 November 2022
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	<math> \text{Use } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math> <br><br>		<math> \text{Use } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math> <br><br><br>

	<math>		<math>

2022-11-29T18:07:58Z

← Older revision		Revision as of 18:07, 29 November 2022
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	<math> \text{~~First use~~ } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math> <br>		<math> \text{Use } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math> <br><br>

	<math>		<math>

2022-11-29T18:07:40Z

← Older revision		Revision as of 18:07, 29 November 2022
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	<math> \text{~~Use~~ } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math~~> <br>~~		<math> \text{First use } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math> <br>
	~~<math>\text{ to evaluate } \int(\ln{x})^3dx</math> <br~~><br>

	<math>		<math>