Difference between revisions of "TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 228"
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<amsmath>\int \frac{\sqrt{x+1}} {x} \cdot dx</amsmath>
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<amsmath>\int \frac{\sqrt{x+1}} {x} \cdot dx </amsmath> <br><br> | <amsmath>\int \frac{\sqrt{x+1}} {x} \cdot dx </amsmath> <br><br> | ||
<amsmath>u = \sqrt{x+1}, u^2 = x+1, u^2 - 1 =x </amsmath> <br><br> | <amsmath>u = \sqrt{x+1}, u^2 = x+1, u^2 - 1 =x </amsmath> <br><br> | ||
| − | <amsmath>dx = 2 \cdot sqrt(x+1) \cdot du = 2 \cdot u \cdot du</amsmath> <br><br> | + | <amsmath>dx = 2 \cdot \sqrt(x+1) \cdot du = 2 \cdot u \cdot du</amsmath> <br><br> |
<amsmath>\int \frac{u} {x} \cdot 2\cdot u du </amsmath> <br><br> | <amsmath>\int \frac{u} {x} \cdot 2\cdot u du </amsmath> <br><br> | ||
<amsmath>2 \cdot \int \frac{1} {u^2-1} du </amsmath> <br><br> | <amsmath>2 \cdot \int \frac{1} {u^2-1} du </amsmath> <br><br> | ||
<amsmath>2 \arccos(u)+c </amsmath> <br><br> | <amsmath>2 \arccos(u)+c </amsmath> <br><br> | ||
<amsmath>2 \arccos(\sqrt{x+1})+c </amsmath> <br><br> | <amsmath>2 \arccos(\sqrt{x+1})+c </amsmath> <br><br> | ||
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== Links == | == Links == | ||
* Diskussion Informatik-Forum [http://www.informatik-forum.at/showthread.php?t=58452 WS07] Beispiel 45 | * Diskussion Informatik-Forum [http://www.informatik-forum.at/showthread.php?t=58452 WS07] Beispiel 45 | ||
* Diskussion Informatik-Forum [http://www.informatik-forum.at/showthread.php?t=63741 SS08] Beispiel 48 | * Diskussion Informatik-Forum [http://www.informatik-forum.at/showthread.php?t=63741 SS08] Beispiel 48 | ||
Revision as of 12:58, 15 July 2008
SS 08 Beispiel 48
Man berechne:
Hilfreiches
Lösungsvorschlag
<amsmath>\int \frac{\sqrt{x+1}} {x} \cdot dx </amsmath>
<amsmath>u = \sqrt{x+1}, u^2 = x+1, u^2 - 1 =x </amsmath>
<amsmath>dx = 2 \cdot \sqrt(x+1) \cdot du = 2 \cdot u \cdot du</amsmath>
<amsmath>\int \frac{u} {x} \cdot 2\cdot u du </amsmath>
<amsmath>2 \cdot \int \frac{1} {u^2-1} du </amsmath>
<amsmath>2 \arccos(u)+c </amsmath>
<amsmath>2 \arccos(\sqrt{x+1})+c </amsmath>