Lösung 2.1:5a - Online Mathematik Brückenkurs 2

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                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
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			Lösung 2.1:5a
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 09:36, 28. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 13:00, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 2:
Zeile 2:
 <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,  <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{\sqrt{x+9}-\sqrt{x}}  \frac{1}{\sqrt{x+9}-\sqrt{x}}
 &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]  &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
Zeile 12:
Zeile 12:
 Thus,  Thus,
  
- {{Displayed math||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
  
 If we write the square roots in power form,  If we write the square roots in power form,
  
- {{Displayed math||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}} + {{Abgesetzte Formel||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
  
 we see that we have a standard integral and can write down the primitive functions directly,  we see that we have a standard integral and can write down the primitive functions directly,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx  \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
 &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]  &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
Zeile 32:
Zeile 32:
 This can also be written with square roots as  This can also be written with square roots as
  
- {{Displayed math||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
  
  
 Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,  Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)  \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
 &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]  &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
 &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}  &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
 \end{align}</math>}}  \end{align}</math>}}
Version vom 13:00, 10. Mär. 2009
If we multiply top and bottom of the fraction by the conjugate expression 
\displaystyle \sqrt{x+9}+\sqrt{x} then the formula for the difference of two squares gives that denominator's root is squared away,





\displaystyle \begin{align}
\frac{1}{\sqrt{x+9}-\sqrt{x}}
&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
\end{align}



Thus,





\displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}


If we write the square roots in power form,





\displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,


we see that we have a standard integral and can write down the primitive functions directly,





\displaystyle \begin{align}
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
\end{align}



where C is an arbitrary constant.
This can also be written with square roots as





\displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}




Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,





\displaystyle \begin{align}
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
\end{align}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:5a“
                       Willkommen zum Kurs
                       
                        Information über den Kurs 
                        Information über Prüfungen
                      
                    
                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
                      Kurs als PDF
                                            
                    
                    
		       Suche
		       
		         

                           
                            
			 
		       
		    
		    
		  
		  
		  
		    
		      
		        
			
			Lösung 2.1:5a
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 09:36, 28. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 13:00, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 2:
Zeile 2:
 <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,  <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{\sqrt{x+9}-\sqrt{x}}  \frac{1}{\sqrt{x+9}-\sqrt{x}}
 &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]  &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
Zeile 12:
Zeile 12:
 Thus,  Thus,
  
- {{Displayed math||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
  
 If we write the square roots in power form,  If we write the square roots in power form,
  
- {{Displayed math||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}} + {{Abgesetzte Formel||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
  
 we see that we have a standard integral and can write down the primitive functions directly,  we see that we have a standard integral and can write down the primitive functions directly,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx  \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
 &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]  &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
Zeile 32:
Zeile 32:
 This can also be written with square roots as  This can also be written with square roots as
  
- {{Displayed math||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
  
  
 Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,  Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)  \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
 &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]  &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
 &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}  &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
 \end{align}</math>}}  \end{align}</math>}}
Version vom 13:00, 10. Mär. 2009
If we multiply top and bottom of the fraction by the conjugate expression 
\displaystyle \sqrt{x+9}+\sqrt{x} then the formula for the difference of two squares gives that denominator's root is squared away,





\displaystyle \begin{align}
\frac{1}{\sqrt{x+9}-\sqrt{x}}
&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
\end{align}



Thus,





\displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}


If we write the square roots in power form,





\displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,


we see that we have a standard integral and can write down the primitive functions directly,





\displaystyle \begin{align}
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
\end{align}



where C is an arbitrary constant.
This can also be written with square roots as





\displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}




Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,





\displaystyle \begin{align}
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
\end{align}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:5a“
-                      Willkommen zum Kurs
                       
                        Information über den Kurs 
                        Information über Prüfungen
                      
                    
                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
                      Kurs als PDF
                                            
                    
                    
		       Suche
+			Lösung 2.1:5a
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 09:36, 28. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 13:00, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 2:
Zeile 2:
 <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,  <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{\sqrt{x+9}-\sqrt{x}}  \frac{1}{\sqrt{x+9}-\sqrt{x}}
 &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]  &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
Zeile 12:
Zeile 12:
 Thus,  Thus,
  
- {{Displayed math||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
  
 If we write the square roots in power form,  If we write the square roots in power form,
  
- {{Displayed math||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}} + {{Abgesetzte Formel||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
  
 we see that we have a standard integral and can write down the primitive functions directly,  we see that we have a standard integral and can write down the primitive functions directly,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx  \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
 &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]  &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
Zeile 32:
Zeile 32:
 This can also be written with square roots as  This can also be written with square roots as
  
- {{Displayed math||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
  
  
 Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,  Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)  \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
 &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]  &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
 &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}  &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
 \end{align}</math>}}  \end{align}</math>}}
Version vom 13:00, 10. Mär. 2009
If we multiply top and bottom of the fraction by the conjugate expression 
\displaystyle \sqrt{x+9}+\sqrt{x} then the formula for the difference of two squares gives that denominator's root is squared away,





\displaystyle \begin{align}
\frac{1}{\sqrt{x+9}-\sqrt{x}}
&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
\end{align}



Thus,





\displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}


If we write the square roots in power form,





\displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,


we see that we have a standard integral and can write down the primitive functions directly,





\displaystyle \begin{align}
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
\end{align}



where C is an arbitrary constant.
This can also be written with square roots as





\displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}




Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,





\displaystyle \begin{align}
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
\end{align}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:5a“
@@ Zeile 2: / Zeile 2: @@
 <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{\sqrt{x+9}-\sqrt{x}}
 &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
@@ Zeile 12: / Zeile 12: @@
 Thus,
-{{Displayed math||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
+{{Abgesetzte Formel||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
 If we write the square roots in power form,
-{{Displayed math||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
+{{Abgesetzte Formel||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
 we see that we have a standard integral and can write down the primitive functions directly,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
 &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]
@@ Zeile 32: / Zeile 32: @@
 This can also be written with square roots as
-{{Displayed math||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
+{{Abgesetzte Formel||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
 Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
 &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]
 &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
 \end{align}</math>}}
 \displaystyle \begin{align}
\frac{1}{\sqrt{x+9}-\sqrt{x}}
&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
\end{align}
 \displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}
 \displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,
 \displaystyle \begin{align}
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
\end{align}
 \displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}
 \displaystyle \begin{align}
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] 
&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
\end{align}