009B Sample Final 3, Problem 6 - Revision history

Kayla Murray: Replaced content with " Find the following integrals (a) $\int \frac{3x-1}{2x^2-x}~dx$ (b) $\int \frac{\sqr..."$

2017-12-03T00:47:56Z

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..."

Show changes

Kayla Murray at 21:50, 12 March 2017

2017-03-12T21:50:24Z

Kayla Murray at 17:14, 3 March 2017

2017-03-03T17:14:10Z

Kayla Murray at 17:07, 3 March 2017

2017-03-03T17:07:03Z

Kayla Murray at 17:00, 3 March 2017

2017-03-03T17:00:22Z

Kayla Murray at 16:56, 3 March 2017

2017-03-03T16:56:54Z

Kayla Murray at 22:59, 2 March 2017

2017-03-02T22:59:35Z

Kayla Murray at 21:14, 2 March 2017

2017-03-02T21:14:44Z

Kayla Murray at 21:10, 2 March 2017

2017-03-02T21:10:48Z

Kayla Murray at 20:46, 2 March 2017

2017-03-02T20:46:17Z

@@ Line 103: / Line 103: @@
 |Hence,
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>dx=2udu.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>dx=2u~du.</math>
 |-
 |Using all this information, we get
@@ Line 163: / Line 163: @@
 |To complete this integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|For the first integral, let &nbsp;<math style="vertical-align: -3px">t=u+1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">dt=du.</math>
+|For the first integral, let &nbsp;<math style="vertical-align: -3px">t=u+1.</math>&nbsp;
 |-
-|For the second integral, let &nbsp;<math style="vertical-align: -2px">v=u-1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">dv=du.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">dt=du.</math>
 |-
 |Finally, we integrate to get

@@ Line 85: / Line 85: @@
 !Step 1: &nbsp;
 |-
-|We begin by using &nbsp;<math>u</math>-substitution.
+|We begin by using &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math>u=\sqrt{x+1}.</math>
+|Let &nbsp;<math style="vertical-align: -3px">u=\sqrt{x+1}.</math>
 |-
-|Then, &nbsp;<math>u^2=x+1</math>&nbsp; and &nbsp;<math>x=u^2-1.</math>
+|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>
 |-
 |Also, we have
@@ Line 101: / Line 101: @@
 \end{array}</math>
 |-
-|Hence, &nbsp;<math>dx=2udu.</math>
+|Hence,
 |-
 |Using all this information, we get
@@ Line 131: / Line 133: @@
 |Now, for the remaining integral, we use partial fraction decomposition.
 |-
-|Let &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.</math>
+|Let
 |-
-|Then, we multiply this equation by &nbsp;<math>(x-1)(x+1)</math>&nbsp; to get
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>2=A(x-1)+B(x+1).</math>
 |-
-|If we let &nbsp;<math>x=1,</math>&nbsp; we get &nbsp;<math>B=1.</math>
+|If we let &nbsp;<math style="vertical-align: -4px">x=1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -2px">B=1.</math>
 |-
-|If we let &nbsp;<math>x=-1,</math>&nbsp; we get &nbsp;<math>A=-1.</math>
+|Thus, we have
 |-
-|Thus, we have &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
 |-
 |Using this equation, we have
@@ Line 155: / Line 161: @@
 !Step 4: &nbsp;
 |-
-|To complete this integral, we need to use &nbsp;<math>u</math>-substitution.
+|To complete this integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|For the first integral, let &nbsp;<math>t=u+1.</math>&nbsp; Then, &nbsp;<math>dt=du.</math>
+|For the first integral, let &nbsp;<math style="vertical-align: -3px">t=u+1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">dt=du.</math>
 |-
-|For the second integral, let &nbsp;<math>v=u-1.</math>&nbsp; Then, &nbsp;<math>dv=du.</math>
+|For the second integral, let &nbsp;<math style="vertical-align: -2px">v=u-1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">dv=du.</math>
 |-
 |Finally, we integrate to get

@@ Line 33: / Line 33: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math>
 |-
-|If we multiply both sides of this equation by &nbsp;<math>x(2x-1),</math>&nbsp; we get
+|If we multiply both sides of this equation by &nbsp;<math style="vertical-align: -5px">x(2x-1),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>3x-1=A(2x-1)+Bx.</math>
@@ Line 41: / Line 41: @@
 !Step 2: &nbsp;
 |-
-|Now, if we let &nbsp;<math>x=0,</math>&nbsp; we get &nbsp;<math>A=1.</math>
+|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>
 |-
-|If we let &nbsp;<math>x=\frac{1}{2},</math>&nbsp; we get &nbsp;<math>B=1.</math>
+|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>
 |-
 |Therefore,
@@ Line 53: / Line 53: @@
 !Step 3: &nbsp;
 |-
-|Therefore, we have
+|Now, we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 63: / Line 63: @@
 \end{array}</math>
 |-
-|Now, we use &nbsp;<math>u</math>-substitution.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math>u=2x-1.</math>
+|Let &nbsp;<math style="vertical-align: -1px">u=2x-1.</math>
 |-
-|Then, &nbsp;<math>du=2dx</math>&nbsp; and &nbsp;<math>\frac{du}{2}=dx.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{2}=dx.</math>
 |-
 |Hence, we have

@@ Line 12: / Line 12: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B.</math>
+|for some constants <math style="vertical-align: -4px">A,B.</math>
 |}
@@ Line 33: / Line 33: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<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
+|If we multiply both sides of this equation by &nbsp;<math>x(2x-1),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>3x-1=A(2x-1)+Bx.</math>
@@ Line 41: / Line 41: @@
 !Step 2: &nbsp;
 |-
-|Now, if we let <math>x=0,</math> we get <math>A=1.</math>
+|Now, if we let &nbsp;<math>x=0,</math>&nbsp; we get &nbsp;<math>A=1.</math>
 |-
-|If we let <math>x=\frac{1}{2},</math> we get <math>B=1.</math>
+|If we let &nbsp;<math>x=\frac{1}{2},</math>&nbsp; we get &nbsp;<math>B=1.</math>
 |-
 |Therefore,
@@ Line 63: / Line 63: @@
 \end{array}</math>
 |-
-|Now, we use <math>u</math>-substitution.
+|Now, we use &nbsp;<math>u</math>-substitution.
 |-
-|Let <math>u=2x-1.</math>
+|Let &nbsp;<math>u=2x-1.</math>
 |-
-|Then, <math>du=2dx</math> and <math>\frac{du}{2}=dx.</math>
+|Then, &nbsp;<math>du=2dx</math>&nbsp; and &nbsp;<math>\frac{du}{2}=dx.</math>
 |-
 |Hence, we have
@@ Line 85: / Line 85: @@
 !Step 1: &nbsp;
 |-
-|We begin by using <math>u</math>-substitution.
+|We begin by using &nbsp;<math>u</math>-substitution.
 |-
-|Let <math>u=\sqrt{x+1}.</math>
+|Let &nbsp;<math>u=\sqrt{x+1}.</math>
 |-
-|Then, <math>u^2=x+1</math> and <math>x=u^2-1.</math>
+|Then, &nbsp;<math>u^2=x+1</math>&nbsp; and &nbsp;<math>x=u^2-1.</math>
 |-
 |Also, we have
@@ Line 101: / Line 101: @@
 \end{array}</math>
 |-
-|Hence, <math>dx=2udu</math>.
+|Hence, &nbsp;<math>dx=2udu.</math>
 |-
 |Using all this information, we get
 |-
-|<math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math>
 |}
@@ Line 131: / Line 131: @@
 |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>
+|Let &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.</math>
 |-
-|Then, we multiply this equation by <math>(x-1)(x+1)</math> to get
+|Then, we multiply this equation by &nbsp;<math>(x-1)(x+1)</math>&nbsp; to get
 |-
-|<math>2=A(x-1)+B(x+1).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>2=A(x-1)+B(x+1).</math>
 |-
-|If we let <math>x=1,</math> we get <math>B=1.</math>
+|If we let &nbsp;<math>x=1,</math>&nbsp; we get &nbsp;<math>B=1.</math>
 |-
-|If we let <math>x=-1,</math> we get <math>A=-1.</math>
+|If we let &nbsp;<math>x=-1,</math>&nbsp; we get &nbsp;<math>A=-1.</math>
 |-
-|Thus, we have <math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
+|Thus, we have &nbsp;<math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
 |-
 |Using this equation, we have
@@ Line 155: / Line 155: @@
 !Step 4: &nbsp;
 |-
-|To complete this integral, we need to use <math>u</math>-substitution.
+|To complete this integral, we need to use &nbsp;<math>u</math>-substitution.
 |-
-|For the first integral, let <math>t=u+1.</math> Then, <math>dt=du.</math>
+|For the first integral, let &nbsp;<math>t=u+1.</math>&nbsp; Then, &nbsp;<math>dt=du.</math>
 |-
-|For the second integral, let <math>v=u-1.</math> Then, <math>dv=du.</math>
+|For the second integral, let &nbsp;<math>v=u-1.</math>&nbsp; Then, &nbsp;<math>dv=du.</math>
 |-
 |Finally, we integrate to get

@@ Line 105: / Line 105: @@
 |Using all this information, we get
 |-
+|<math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math>
 |}
@@ Line 111: / Line 111: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 122: / Line 180: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>\ln |x|+\frac{1}{2}\ln |2x-1|+C</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C</math>
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 |First, we factor the denominator to get
 |-
-|<math>\int \frac{3x-1}{2x^2-x}~dx=\int \frac{3x-1}{x(2x-1)}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<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.
@@ Line 31: / Line 31: @@
 |We let
 |-
-|<math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math>
 |}