Difference between revisions of "TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 228"

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{{Substitutionsregel}}
 
{{Substitutionsregel}}
  
==Lösungsvorschlag==
+
==Lösungsvorschlag von Ven==
 
<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>

Revision as of 12:59, 15 July 2008

SS 08 Beispiel 48

Man berechne:

<amsmath>\int \frac{\sqrt{x+1}} {x} \cdot dx</amsmath>

Hilfreiches

Vorlage:Substitutionsregel

Lösungsvorschlag von Ven

<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>

Links

  • Diskussion Informatik-Forum WS07 Beispiel 45
  • Diskussion Informatik-Forum SS08 Beispiel 48