7.1 Integration By Parts/37 - Revision history

Elizabethh58413 at 03:09, 29 November 2022

2022-11-29T03:09:35Z

← Older revision		Revision as of 03:09, 29 November 2022
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	\end{align}		\end{align}
	</math> <br>		</math> <br>
			<math>\int(u-1)\ln({u}) du </math> <br><br>
			<math> w = \ln{u} \qquad dv = (u-1)dv </math> <br><br>
			<math> du = \frac{1}{u}du \qquad v = \frac{1}{2}u^2-u </math> <br><br>
	<math>		<math>
	\begin{align}		\begin{align}

Elizabethh58413 at 02:58, 29 November 2022

2022-11-29T02:58:28Z

← Older revision		Revision as of 02:58, 29 November 2022
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	&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex]		&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex]
	\end{align}		\end{align}
	<\math>		</math>

Elizabethh58413 at 02:57, 29 November 2022

2022-11-29T02:57:44Z

@@ Line 9: / Line 9: @@
 <math>
 \begin{align}
-\int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]
+\int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]

Elizabethh58413: Created page with " $\int x\ln({x})dx$
$\begin{align} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align}$
$\begin{align} \int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]"$

2022-11-29T02:37:22Z

Created page with "<math> \int x\ln({x})dx </math> <br> <math> \begin{align} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align} </math> <br> <math> \begin{align} \int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]"

New page

<math> \int x\ln({x})dx </math> <br>
<math>
\begin{align}
u &= x + 1 \\[2ex]
x &= u-1 \\[2ex]
du &= dx \\[2ex]
\end{align}
</math> <br>
<math>
\begin{align}
\int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]

7.1 Integration By Parts/37 - Revision history

Elizabethh58413 at 03:09, 29 November 2022

Elizabethh58413 at 02:58, 29 November 2022

Elizabethh58413 at 02:57, 29 November 2022

Elizabethh58413: Created page with " \int x\ln({x})dx \begin{align} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align} \begin{align} \int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]"

Elizabethh58413: Created page with " $\int x\ln({x})dx$
$\begin{align} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align}$
$\begin{align} \int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]"$