MathAdmin: Created page with " Find the following integrals (a) $\int \frac{3x-1}{2x^2-x}~dx$ (b) $\int \frac{\sqrt{x+..."$

2017-04-10T16:55:00Z

Created page with " Find the following integrals (a) <math>\int \frac{3x-1}{2x^2-x}~dx</math> (b) <math>\int \frac{\sqrt{x+..."

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Find the following integrals

(a)  <math>\int \frac{3x-1}{2x^2-x}~dx</math>

(b)  <math>\int \frac{\sqrt{x+1}}{x}~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Through partial fraction decomposition, we can write the fraction
|-
|       <math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
|-
|for some constants <math style="vertical-align: -4px">A,B.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we factor the denominator to get
|-
|       <math>\int \frac{3x-1}{2x^2-x}~dx=\int \frac{3x-1}{x(2x-1)}.</math>
|-
|We use the method of partial fraction decomposition.
|-
|We let
|-
|       <math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math>
|-
|If we multiply both sides of this equation by  <math style="vertical-align: -5px">x(2x-1),</math>  we get
|-
|       <math>3x-1=A(2x-1)+Bx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, if we let  <math style="vertical-align: -4px">x=0,</math>  we get  <math style="vertical-align: -2px">A=1.</math>
|-
|If we let  <math style="vertical-align: -13px">x=\frac{1}{2},</math>  we get  <math style="vertical-align: -2px">B=1.</math>
|-
|Therefore,
|-
|       <math>\frac{3x-1}{x(2x-1)}=\frac{1}{x}+\frac{1}{2x-1}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\int \frac{1}{x}+\frac{1}{2x-1}~dx}\\
&&\\
& = & \displaystyle{\int \frac{1}{x}~dx+\int \frac{1}{2x-1}~dx}\\
&&\\
& = & \displaystyle{\ln |x|+\int \frac{1}{2x-1}~dx.}
\end{array}</math>
|-
|Now, we use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let  <math style="vertical-align: -1px">u=2x-1.</math>
|-
|Then,  <math style="vertical-align: -1px">du=2dx</math>  and  <math style="vertical-align: -13px">\frac{du}{2}=dx.</math>
|-
|Hence, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\ln |x|+\frac{1}{2}\int \frac{1}{u}~du}\\
&&\\
& = & \displaystyle{\ln |x|+\frac{1}{2}\ln |u|+C}\\
&&\\
& = & \displaystyle{\ln |x|+\frac{1}{2}\ln |2x-1|+C.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using  <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let  <math style="vertical-align: -3px">u=\sqrt{x+1}.</math>
|-
|Then,  <math style="vertical-align: -2px">u^2=x+1</math>  and  <math style="vertical-align: -1px">x=u^2-1.</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{du} & = & \displaystyle{\frac{1}{2} (x+1)^{\frac{-1}{2}}dx}\\
&&\\
& = & \displaystyle{\frac{1}{2\sqrt{x+1}}dx}\\
&&\\
& = & \displaystyle{\frac{1}{2u}dx.}
\end{array}</math>
|-
|Hence,
|-
|       <math>dx=2u~du.</math>
|-
|Using all this information, we get
|-
|       <math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{\int \frac{2u^2-2+2}{u^2-1}~du}\\
&&\\
& = & \displaystyle{\int \frac{2(u^2-1)}{u^2-1}~du+\int \frac{2}{u^2-1}~du}\\
&&\\
& = & \displaystyle{\int 2~du+\int \frac{2}{u^2-1}~du}\\
&&\\
& = & \displaystyle{2u+\int \frac{2}{u^2-1}~du}\\
&&\\
& = & \displaystyle{2\sqrt{x+1}+\int \frac{2}{(u-1)(u+1)}~du.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, for the remaining integral, we use partial fraction decomposition.
|-
|Let
|-
|       <math>\frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.</math>
|-
|Then, we multiply this equation by  <math style="vertical-align: -5px">(x-1)(x+1)</math>  to get
|-
|       <math>2=A(x-1)+B(x+1).</math>
|-
|If we let  <math style="vertical-align: -4px">x=1,</math>  we get  <math style="vertical-align: -2px">B=1.</math>
|-
|If we let  <math style="vertical-align: -4px">x=-1,</math>  we get  <math style="vertical-align: -2px">A=-1.</math>
|-
|Thus, we have
|-
|       <math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
|-
|Using this equation, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{(u+1)}+\frac{1}{u-1}~du}\\
&&\\
& = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{(u+1)}~du+\int \frac{1}{u-1}~du.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|To complete this integral, we need to use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|For the first integral, let  <math style="vertical-align: -3px">t=u+1.</math> 
|-
|Then,  <math style="vertical-align: -1px">dt=du.</math>
|-
|For the second integral, let  <math style="vertical-align: -2px">v=u-1.</math> 
|-
|Then,  <math style="vertical-align: -1px">dv=du.</math>
|-
|Finally, we integrate to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{t}~dt+\int \frac{1}{v}~dv}\\
&&\\
& = & \displaystyle{2\sqrt{x+1}+\ln|t|+\ln|v|+C}\\
&&\\
& = & \displaystyle{2\sqrt{x+1}+\ln|u+1|+\ln|u-1|+C}\\
&&\\
& = & \displaystyle{2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''   <math>\ln |x|+\frac{1}{2}\ln |2x-1|+C</math>
|-
|   '''(b)'''   <math>2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C</math>
|}
[[009B_Sample_Final_3|'''Return to Sample Exam''']]

009B Sample Final 3, Problem 6 - Revision history

MathAdmin: Created page with " Find the following integrals (a) \int \frac{3x-1}{2x^2-x}~dx (b) \int \frac{\sqrt{x+..."

MathAdmin: Created page with " Find the following integrals (a) $\int \frac{3x-1}{2x^2-x}~dx$ (b) $\int \frac{\sqrt{x+..."$