5.5 The Substitution Rule/19 - Revision history

Josuem95981@students.laalliance.org at 06:59, 3 September 2022

2022-09-03T06:59:04Z

← Older revision		Revision as of 06:59, 3 September 2022
Line 5:		Line 5:
	& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]		& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]

	& u=\ln(x)//		& u=\ln(x) \\
	& du=\frac{1}{x}dx//		& du=\frac{1}{x}dx \\

Josuem95981@students.laalliance.org at 06:58, 3 September 2022

2022-09-03T06:58:46Z

← Older revision		Revision as of 06:58, 3 September 2022
Line 5:		Line 5:
	& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]		& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]

	& u=\ln(x)		& u=\ln(x)//
	& du=\frac{1}{x}dx		& du=\frac{1}{x}dx//

Josuem95981@students.laalliance.org at 06:58, 3 September 2022

2022-09-03T06:58:18Z

← Older revision		Revision as of 06:58, 3 September 2022
Line 5:		Line 5:
	& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]		& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]

	u=\ln(x)		& u=\ln(x)
	du=\frac{1}{x}dx		& du=\frac{1}{x}dx
	~~du=\frac{dx}{x}~~

	& = \ \frac{1}{3}(\ln(x))^3+C		& = \ \frac{1}{3}(\ln(x))^3+C

Josuem95981@students.laalliance.org at 06:55, 3 September 2022

2022-09-03T06:55:48Z

← Older revision		Revision as of 06:55, 3 September 2022
Line 7:		Line 7:
	u=\ln(x)		u=\ln(x)
	du=\frac{1}{x}dx		du=\frac{1}{x}dx
	du=frac{dx}{x}		du=\frac{dx}{x}

	& = \ \frac{1}{3}(\ln(x))^3+C		& = \ \frac{1}{3}(\ln(x))^3+C

Josuem95981@students.laalliance.org at 06:55, 3 September 2022

2022-09-03T06:55:37Z

← Older revision		Revision as of 06:55, 3 September 2022
Line 5:		Line 5:
	& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]		& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]

			u=\ln(x)
			du=\frac{1}{x}dx
			du=frac{dx}{x}

	& = \ \frac{1}{3}(\ln(x))^3+C		& = \ \frac{1}{3}(\ln(x))^3+C

Josuem95981@students.laalliance.org at 06:54, 3 September 2022

2022-09-03T06:54:01Z

← Older revision		Revision as of 06:54, 3 September 2022
Line 5:		Line 5:
	& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]		& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]

	~~----~~

	& = \ \frac{1}{3}(\ln(x))^3+C		& = \ \frac{1}{3}(\ln(x))^3+C

Josuem95981@students.laalliance.org at 06:53, 3 September 2022

2022-09-03T06:53:51Z

← Older revision		Revision as of 06:53, 3 September 2022
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	\begin{align}		\begin{align}

	\int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du		& \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du \\[2ex]
	= \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C		& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]

	----		----

	= \ \frac{1}{3}(\ln(x))^3+C		& = \ \frac{1}{3}(\ln(x))^3+C

Josuem95981@students.laalliance.org: Created page with " $\begin{align} \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C ---- = \ \frac{1}{3}(\ln(x))^3+C \end{align}$ "

2022-09-03T06:51:49Z

Created page with "<math> \begin{align} \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C ---- = \ \frac{1}{3}(\ln(x))^3+C \end{align} </math>"

New page

<math>
\begin{align}

\int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du
= \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C

----

= \ \frac{1}{3}(\ln(x))^3+C

\end{align}
</math>

5.5 The Substitution Rule/19 - Revision history

Josuem95981@students.laalliance.org at 06:59, 3 September 2022

Josuem95981@students.laalliance.org at 06:58, 3 September 2022

Josuem95981@students.laalliance.org at 06:58, 3 September 2022

Josuem95981@students.laalliance.org at 06:55, 3 September 2022

Josuem95981@students.laalliance.org at 06:55, 3 September 2022

Josuem95981@students.laalliance.org at 06:54, 3 September 2022

Josuem95981@students.laalliance.org at 06:53, 3 September 2022

Josuem95981@students.laalliance.org: Created page with " \begin{align} \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C ---- = \ \frac{1}{3}(\ln(x))^3+C \end{align} "

Josuem95981@students.laalliance.org: Created page with " $\begin{align} \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C ---- = \ \frac{1}{3}(\ln(x))^3+C \end{align}$ "