|
|
| (5 intermediate revisions by the same user not shown) |
| Line 5: |
Line 5: |
| | <span class="exam">(b) <math>\int \frac{\sqrt{x+1}}{x}~dx</math> | | <span class="exam">(b) <math>\int \frac{\sqrt{x+1}}{x}~dx</math> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |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:''' | + | [[009B Sample Final 3, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''(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>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>x=0,</math> we get <math>A=1.</math>
| |
| − | |-
| |
| − | |If we let <math>x=\frac{1}{2},</math> we get <math>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:
| |
| − | |-
| |
| − | |Therefore, 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>u</math>-substitution.
| |
| − | |-
| |
| − | |Let <math>u=2x-1.</math>
| |
| − | |-
| |
| − | |Then, <math>du=2dx</math> and <math>\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>u</math>-substitution.
| |
| − | |-
| |
| − | |Let <math>u=\sqrt{x+1}.</math>
| |
| − | |-
| |
| − | |Then, <math>u^2=x+1</math> and <math>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=2udu</math>.
| |
| − | |-
| |
| − | |Using all this information, we get
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | '''(a)''' <math>\ln |x|+\frac{1}{2}\ln |2x-1|+C</math>
| |
| − | |-
| |
| − | |'''(b)'''
| |
| − | |}
| |
| | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
