Lösung 2.2:3e - Versionsgeschichte

Sekretariat1 um 12:45, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 12:45:06 GMT

← Nächstältere Version		Version vom 12:45, 27. Aug. 2009
Zeile 25:		Zeile 25:
	\end{align}</math>}}		\end{align}</math>}}

-	Im letzten Schritt haben wir das Betragszeichen weggelassen, da <math>x^2+1</math>immer positiv ist.	+	Im letzten Schritt haben wir das Betragszeichen weggelassen, da <math>x^2+1</math> immer positiv ist.

Sekretariat1 um 12:44, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 12:44:45 GMT

Martin um 13:29, 5. Mai 2009

Martin — Tue, 05 May 2009 13:29:31 GMT

← Nächstältere Version		Version vom 13:29, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~If we differentiate the denominator in the integrand~~	+	Die Ableitung des Nenners ist

	{{Abgesetzte Formel\|\|<math>(x^2+1)' = 2x</math>}}		{{Abgesetzte Formel\|\|<math>(x^2+1)' = 2x</math>}}

-	~~we obtain almost the same expression as in the numerator; there is a constant 2 which is different~~. ~~We therefore rewrite the numerator as~~	+	Dies ist fast der Zähler. Wir schreiben den Zähler wie

	{{Abgesetzte Formel\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}		{{Abgesetzte Formel\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}

-	~~so the integral can be written as~~	+	sodass wir das folgende Integral erhalten,

	{{Abgesetzte Formel\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}

-	~~and we see that the substitution~~ <math>u=x^2+1</math> ~~can be used to simplify the integral~~,	+	Wir sehen hier dass die Substitution <math>u=x^2+1</math> günstig ist,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 25:		Zeile 25:
	\end{align}</math>}}		\end{align}</math>}}

-	~~In the last step~~, ~~we take away the absolute sign around the argument in <math>\ln</math>, because~~ <math>x^2+1</math> ~~is always greater than or equal to 1~~.	+	Im letzten Schritt haben wir das Betragszeichen weggelassen, nachdem <math>x^2+1</math>immer positiv ist.

Tekbot: Solution 2.2:3e moved to Lösung 2.2:3e: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:23:55 GMT

Solution 2.2:3e moved to Lösung 2.2:3e: Robot: moved page

← Nächstältere Version

Version vom 10:23, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Tue, 10 Mar 2009 13:02:24 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 13:02, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we differentiate the denominator in the integrand		If we differentiate the denominator in the integrand

-	{{~~Displayed math~~\|\|<math>(x^2+1)' = 2x</math>}}	+	{{Abgesetzte Formel\|\|<math>(x^2+1)' = 2x</math>}}

	we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as		we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as

-	{{~~Displayed math~~\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}	+	{{Abgesetzte Formel\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}

	so the integral can be written as		so the integral can be written as

-	{{~~Displayed math~~\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}

	and we see that the substitution <math>u=x^2+1</math> can be used to simplify the integral,		and we see that the substitution <math>u=x^2+1</math> can be used to simplify the integral,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int \frac{3x}{x^2+1}\,dx		\int \frac{3x}{x^2+1}\,dx
	&= \left\{ \begin{align}		&= \left\{ \begin{align}

Tek um 14:53, 28. Okt. 2008

Tek — Tue, 28 Oct 2008 14:53:13 GMT

← Nächstältere Version		Version vom 14:53, 28. Okt. 2008
Zeile 1:		Zeile 1:
	If we differentiate the denominator in the integrand		If we differentiate the denominator in the integrand

		+	{{Displayed math\|\|<math>(x^2+1)' = 2x</math>}}

-	~~<math>3x=\frac{3}{~~2~~}\centerdot 2x=\frac{3}{2}\left( x^{2}+1 \right)^{\prime }</math>~~	+	we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as

-		+	{{Displayed math\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}
-	~~we obtain almost the same expression as in the numerator; there is a constant~~	+
-	~~<math>\text~~{~~2}</~~math>	+
-	~~which is different. We therefore rewrite the numerator as~~	+
-		+
-		+
-	<math>3x=\frac{3}{2}\~~centerdot~~ 2x=\frac{3}{2}\~~centerdot \left~~( x^{2}+1 ~~\right~~)~~^{\prime }~~</math>,	+

	so the integral can be written as		so the integral can be written as

		+	{{Displayed math\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}

-	~~<math>\int{\frac{\frac{3}{2}}{x^{2}+1}}\centerdot \left( x^{2}+1 \right)^{\prime }\,dx</math>,~~	+	and we see that the substitution <math>u=x^2+1</math> can be used to simplify the integral,
-		+
-	and we see that the substitution	+
-	<math>u=x^{2}+1</math>	+
-	can be used to simplify the integral:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \int{\frac{3x}{x^{2}+1}}\~~,~~dx=\left\{ \begin{matrix}~~	+
-	~~u=x^{2}+1 \\~~	+
-	~~du=\left( x^{2}+1 \right)^{\prime }\,dx=2x\,dx \\~~	+
-	~~\end{matrix} \right\} \\~~	+
-	~~& =\frac{3}{2}\int{\frac{\,du}{u}}=\frac{3}{2}\ln \left\| u \right\|+C \\~~	+
-	~~& =\frac{3}{2}\ln \left\| x^{2}+1 \right\|+C \\~~	+
-	~~& =\frac{3}{2}\ln \left( x^{2}+1 \right)+C \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int \frac{3x}{x^2+1}\,dx
		+	&= \left\{ \begin{align}
		+	u &= x^2+1\\[5pt]
		+	du &= (x^2+1)'\,dx = 2x\,dx
		+	\end{align}\right\}\\[5pt]
		+	&= \frac{3}{2}\int \frac{du}{u}\\[5pt]
		+	&= \frac{3}{2}\ln \|u\|+C\\[5pt]
		+	&= \frac{3}{2}\ln \|x^{2}+1\|+C\\[5pt]
		+	&= \frac{3}{2}\ln (x^{2}+1) + C\,\textrm{.}
		+	\end{align}</math>}}

-	In the last step, we take away the absolute sign around the argument in	+	In the last step, we take away the absolute sign around the argument in <math>\ln</math>, because <math>x^2+1</math> is always greater than or equal to 1.
-	<math>\ln </math>, because	+
-	<math>x^{2}+1</math>	+
-	is always greater than or equal to	+
-	~~<math>\text{~~1~~}</math>~~.	+

Ian um 11:16, 21. Okt. 2008

Ian — Tue, 21 Oct 2008 11:16:55 GMT

← Nächstältere Version		Version vom 11:16, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	If we differentiate the denominator in the integrand
-	<~~center~~> ~~[[Image:2_2_3e~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\left( x^{2}+1 \right)^{\prime }</math>
		+
		+
		+	we obtain almost the same expression as in the numerator; there is a constant
		+	<math>\text{2}</math>
		+	which is different. We therefore rewrite the numerator as
		+
		+
		+	<math>3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\centerdot \left( x^{2}+1 \right)^{\prime }</math>,
		+
		+	so the integral can be written as
		+
		+
		+	<math>\int{\frac{\frac{3}{2}}{x^{2}+1}}\centerdot \left( x^{2}+1 \right)^{\prime }\,dx</math>,
		+
		+	and we see that the substitution
		+	<math>u=x^{2}+1</math>
		+	can be used to simplify the integral:
		+
		+
		+	<math>\begin{align}
		+	& \int{\frac{3x}{x^{2}+1}}\,dx=\left\{ \begin{matrix}
		+	u=x^{2}+1 \\
		+	du=\left( x^{2}+1 \right)^{\prime }\,dx=2x\,dx \\
		+	\end{matrix} \right\} \\
		+	& =\frac{3}{2}\int{\frac{\,du}{u}}=\frac{3}{2}\ln \left\| u \right\|+C \\
		+	& =\frac{3}{2}\ln \left\| x^{2}+1 \right\|+C \\
		+	& =\frac{3}{2}\ln \left( x^{2}+1 \right)+C \\
		+	\end{align}</math>
		+
		+
		+	In the last step, we take away the absolute sign around the argument in
		+	<math>\ln </math>, because
		+	<math>x^{2}+1</math>
		+	is always greater than or equal to
		+	<math>\text{1}</math>.

Tekbot: Lösning 2.2:3e moved to Solution 2.2:3e: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:44:21 GMT

Lösning 2.2:3e moved to Solution 2.2:3e: Robot: moved page

← Nächstältere Version

Version vom 14:44, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 08:01:54 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

← Nächstältere Version		Version vom 08:01, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_3e.gif]] </center>	+	<center> [[Image:2_2_3e.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_2_3e.gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 09:30:46 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_3e.gif </center> {{NAVCONTENT_STOP}}

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← Nächstältere Version		Version vom 12:44, 27. Aug. 2009
Zeile 1:		Zeile 1:
	Die Ableitung des Nenners ist		Die Ableitung des Nenners ist

-	{{Abgesetzte Formel\|\|<math>(x^2+1)' = 2x</math>}}	+	{{Abgesetzte Formel\|\|<math>(x^2+1)' = 2x</math>.}}

-	Dies ~~ist~~ fast ~~der~~ Zähler. Wir schreiben den Zähler wie	+	Dies entspricht fast dem Zähler. Wir schreiben den Zähler wie

	{{Abgesetzte Formel\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}		{{Abgesetzte Formel\|\|<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}

-	sodass wir das folgende Integral erhalten,	+	sodass wir das folgende Integral erhalten

-	{{Abgesetzte Formel\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,</math>.}}

-	Wir sehen hier dass die Substitution <math>u=x^2+1</math> günstig ist,	+	Wir sehen hier, dass die Substitution <math>u=x^2+1</math> günstig ist.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 22:		Zeile 22:
	&= \frac{3}{2}\ln \|u\|+C\\[5pt]		&= \frac{3}{2}\ln \|u\|+C\\[5pt]
	&= \frac{3}{2}\ln \|x^{2}+1\|+C\\[5pt]		&= \frac{3}{2}\ln \|x^{2}+1\|+C\\[5pt]
-	&= \frac{3}{2}\ln (x^{2}+1) + C~~\,\textrm{.}~~	+	&= \frac{3}{2}\ln (x^{2}+1) + C
	\end{align}</math>}}		\end{align}</math>}}

-	Im letzten Schritt haben wir das Betragszeichen weggelassen, ~~nachdem~~ <math>x^2+1</math>immer positiv ist.	+	Im letzten Schritt haben wir das Betragszeichen weggelassen, da <math>x^2+1</math>immer positiv ist.