Difference between revisions of "009B Sample Final 3, Problem 6"

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(Replaced content with "<span class="exam"> Find the following integrals <span class="exam">(a)  <math>\int \frac{3x-1}{2x^2-x}~dx</math> <span class="exam">(b)  <math>\int \frac{\sqr...")
 
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<span class="exam">(b) &nbsp;<math>\int \frac{\sqrt{x+1}}{x}~dx</math>
 
<span class="exam">(b) &nbsp;<math>\int \frac{\sqrt{x+1}}{x}~dx</math>
  
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<hr>
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!Foundations: &nbsp;
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[[009B Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']]
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|-
 
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|Through partial fraction decomposition, we can write the fraction
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
 
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|for some constants <math style="vertical-align: -4px">A,B.</math>
 
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|}
 
  
  
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'''Solution:'''
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[[009B Sample Final 3, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''(a)'''
 
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|First, we factor the denominator to get
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{3x-1}{2x^2-x}~dx=\int \frac{3x-1}{x(2x-1)}.</math>
 
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|-
 
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|We use the method of partial fraction decomposition.
 
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|-
 
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|We let
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math>
 
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|-
 
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|If we multiply both sides of this equation by &nbsp;<math style="vertical-align: -5px">x(2x-1),</math>&nbsp; we get
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>3x-1=A(2x-1)+Bx.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Now, if we let &nbsp;<math style="vertical-align: -4px">x=0,</math>&nbsp; we get &nbsp;<math style="vertical-align: -2px">A=1.</math>
 
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|-
 
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|If we let &nbsp;<math style="vertical-align: -13px">x=\frac{1}{2},</math>&nbsp; we get &nbsp;<math style="vertical-align: -2px">B=1.</math>
 
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|-
 
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|Therefore,
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{3x-1}{x(2x-1)}=\frac{1}{x}+\frac{1}{2x-1}.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|Now, we have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\int \frac{1}{x}+\frac{1}{2x-1}~dx}\\
 
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&&\\
 
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& = & \displaystyle{\int \frac{1}{x}~dx+\int \frac{1}{2x-1}~dx}\\
 
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&&\\
 
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& = & \displaystyle{\ln |x|+\int \frac{1}{2x-1}~dx.}
 
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\end{array}</math>
 
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|-
 
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|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -1px">u=2x-1.</math>
 
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|-
 
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|Then, &nbsp;<math style="vertical-align: -1px">du=2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{2}=dx.</math>
 
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|-
 
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|Hence, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\ln |x|+\frac{1}{2}\int \frac{1}{u}~du}\\
 
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&&\\
 
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& = & \displaystyle{\ln |x|+\frac{1}{2}\ln |u|+C}\\
 
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&&\\
 
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& = & \displaystyle{\ln |x|+\frac{1}{2}\ln |2x-1|+C.}
 
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\end{array}</math>
 
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|}
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|We begin by using &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -3px">u=\sqrt{x+1}.</math>
 
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|-
 
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|Then, &nbsp;<math style="vertical-align: -2px">u^2=x+1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=u^2-1.</math>
 
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|-
 
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|Also, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{du} & = & \displaystyle{\frac{1}{2} (x+1)^{\frac{-1}{2}}dx}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{2\sqrt{x+1}}dx}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{2u}dx.}
 
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\end{array}</math>
 
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|-
 
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|Hence,
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>dx=2u~du.</math>
 
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|Using all this information, we get
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Now, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{\int \frac{2u^2-2+2}{u^2-1}~du}\\
 
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&&\\
 
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& = & \displaystyle{\int \frac{2(u^2-1)}{u^2-1}~du+\int \frac{2}{u^2-1}~du}\\
 
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&&\\
 
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& = & \displaystyle{\int 2~du+\int \frac{2}{u^2-1}~du}\\
 
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&&\\
 
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& = & \displaystyle{2u+\int \frac{2}{u^2-1}~du}\\
 
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&&\\
 
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& = & \displaystyle{2\sqrt{x+1}+\int \frac{2}{(u-1)(u+1)}~du.}
 
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\end{array}</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|Now, for the remaining integral, we use partial fraction decomposition.
 
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|-
 
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|Let
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.</math>
 
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|-
 
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|Then, we multiply this equation by &nbsp;<math style="vertical-align: -5px">(x-1)(x+1)</math>&nbsp; to get
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>2=A(x-1)+B(x+1).</math>
 
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|If we let &nbsp;<math style="vertical-align: -4px">x=1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -2px">B=1.</math>
 
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|If we let &nbsp;<math style="vertical-align: -4px">x=-1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -2px">A=-1.</math>
 
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|Thus, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
 
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|Using this equation, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{(u+1)}+\frac{1}{u-1}~du}\\
 
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&&\\
 
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& = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{(u+1)}~du+\int \frac{1}{u-1}~du.}\\
 
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\end{array}</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 4: &nbsp;
 
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|-
 
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|To complete this integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
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|For the first integral, let &nbsp;<math style="vertical-align: -3px">t=u+1.</math>&nbsp;
 
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|Then, &nbsp;<math style="vertical-align: -1px">dt=du.</math>
 
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|For the second integral, let &nbsp;<math style="vertical-align: -2px">v=u-1.</math>&nbsp;
 
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|Then, &nbsp;<math style="vertical-align: -1px">dv=du.</math>
 
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|Finally, we integrate to get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{t}~dt+\int \frac{1}{v}~dv}\\
 
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&&\\
 
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& = & \displaystyle{2\sqrt{x+1}+\ln|t|+\ln|v|+C}\\
 
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&&\\
 
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& = & \displaystyle{2\sqrt{x+1}+\ln|u+1|+\ln|u-1|+C}\\
 
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&&\\
 
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& = & \displaystyle{2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C.}
 
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\end{array}</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>\ln |x|+\frac{1}{2}\ln |2x-1|+C</math>
 
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C</math>
 
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|}
 
 
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:47, 2 December 2017

Find the following integrals

(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{3x-1}{2x^2-x}~dx}

(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{\sqrt{x+1}}{x}~dx}

Solution


Detailed Solution


Return to Sample Exam

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