Solution 2.2:4b - Förberedande kurs i matematik 2

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                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
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+			Solution 2.2:4b
			
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- {{NAVCONTENT_START}} + We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator.  Instead, we take out a factor 3 from the denominator,
- <center> [[Image:2_2_4b-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\begin{align}
- {{NAVCONTENT_START}} + \int \frac{dx}{(x-1)^2+3}
- <center> [[Image:2_2_4b-2(2).gif]] </center> + &= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt] 
- {{NAVCONTENT_STOP}} + &= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} 
  + \end{align}</math>}}
  +  
  + and move a factor <math>\tfrac{1}{3}</math> into the square  <math>(x-1)^2</math>,
  +  
  + {{Displayed math||<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}
  +  
  + Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
  + &= \left\{\begin{align}
  + u &= (x-1)/\!\sqrt{3}\\[5pt]
  + du &= dx/\!\sqrt{3}
  + \end{align}\right\}\\[5pt] 
  + &= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt]
  + &= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt] 
  + &= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt] 
  + &= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.}
  + \end{align}</math>}}
Current revision
We could substitute \displaystyle u=x-1, but we would then still have the problem of the second term, 3, in the denominator.  Instead, we take out a factor 3 from the denominator,





\displaystyle \begin{align}
\int \frac{dx}{(x-1)^2+3}
&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt] 
&= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} 
\end{align}



and move a factor \displaystyle \tfrac{1}{3} into the square  \displaystyle (x-1)^2,





\displaystyle \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}


Now, we substitute \displaystyle u = (x-1)/\!\sqrt{3} and get rid of all the problems at once,





\displaystyle \begin{align}
\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
&= \left\{\begin{align}
u &= (x-1)/\!\sqrt{3}\\[5pt]
du &= dx/\!\sqrt{3}
\end{align}\right\}\\[5pt] 
&= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt]
&= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt] 
&= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt] 
&= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.}
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_2.2:4b"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator.  Instead, we take out a factor 3 from the denominator,
-<center> [[Image:2_2_4b-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
-{{NAVCONTENT_START}}
+\int \frac{dx}{(x-1)^2+3}
-<center> [[Image:2_2_4b-2(2).gif]] </center>
+&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]
-{{NAVCONTENT_STOP}}
+&= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1}
+\end{align}</math>}}
+and move a factor <math>\tfrac{1}{3}</math> into the square  <math>(x-1)^2</math>,
+{{Displayed math||<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}
+Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,
+{{Displayed math||<math>\begin{align}
+\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
+&= \left\{\begin{align}
+u &= (x-1)/\!\sqrt{3}\\[5pt]
+du &= dx/\!\sqrt{3}
+\end{align}\right\}\\[5pt]
+&= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt]
+&= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt]
+&= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt]
+&= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
\int \frac{dx}{(x-1)^2+3}
&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt] 
&= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} 
\end{align}
 \displaystyle \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}
 \displaystyle \begin{align}
\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
&= \left\{\begin{align}
u &= (x-1)/\!\sqrt{3}\\[5pt]
du &= dx/\!\sqrt{3}
\end{align}\right\}\\[5pt] 
&= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt]
&= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt] 
&= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt] 
&= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.}
\end{align}