Difference between revisions of "009B Sample Final 3, Problem 2"

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!Foundations:    
 
!Foundations:    
 
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|<math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math>
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|'''1.'''
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math>
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|-
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|'''2.''' How would you integrate &nbsp; <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math>
 
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|-
 
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|
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&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\ln(x).</math>
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
 
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|-
 
|
 
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&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 
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|-
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
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&&\\
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& = & \displaystyle{\frac{u^2}{2}+C}\\
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&&\\
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& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
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\end{array}</math>
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|}
 
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Revision as of 16:07, 1 March 2017

Evaluate the following integrals.

(a)  

(b)  

(c)  

Foundations:  
1.
       
2. How would you integrate  

        You could use  -substitution.

        Let  
        Then,  

        Thus,

       


Solution:

(a)

Step 1:  
Step 2:  

(b)

Step 1:  
Step 2:  

(c)

Step 1:  
Step 2:  


Final Answer:  
(a)
(b)
(c)

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