Difference between revisions of "009B Sample Final 3, Problem 6"
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| <math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math> | | <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> |
|} | |} | ||
| Line 33: | Line 33: | ||
| <math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math> | | <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 <math>x(2x-1),</math> we get |
|- | |- | ||
| <math>3x-1=A(2x-1)+Bx.</math> | | <math>3x-1=A(2x-1)+Bx.</math> | ||
| Line 41: | Line 41: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |Now, if we let <math>x=0,</math> we get <math>A=1.</math> | + | |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> | + | |If we let <math>x=\frac{1}{2},</math> we get <math>B=1.</math> |
|- | |- | ||
|Therefore, | |Therefore, | ||
| Line 63: | Line 63: | ||
\end{array}</math> | \end{array}</math> | ||
|- | |- | ||
| − | |Now, we use <math>u</math>-substitution. | + | |Now, we use <math>u</math>-substitution. |
|- | |- | ||
| − | |Let <math>u=2x-1.</math> | + | |Let <math>u=2x-1.</math> |
|- | |- | ||
| − | |Then, <math>du=2dx</math> and <math>\frac{du}{2}=dx.</math> | + | |Then, <math>du=2dx</math> and <math>\frac{du}{2}=dx.</math> |
|- | |- | ||
|Hence, we have | |Hence, we have | ||
| Line 85: | Line 85: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | |We begin by using <math>u</math>-substitution. | + | |We begin by using <math>u</math>-substitution. |
|- | |- | ||
| − | |Let <math>u=\sqrt{x+1}.</math> | + | |Let <math>u=\sqrt{x+1}.</math> |
|- | |- | ||
| − | |Then, <math>u^2=x+1</math> and <math>x=u^2-1.</math> | + | |Then, <math>u^2=x+1</math> and <math>x=u^2-1.</math> |
|- | |- | ||
|Also, we have | |Also, we have | ||
| Line 101: | Line 101: | ||
\end{array}</math> | \end{array}</math> | ||
|- | |- | ||
| − | |Hence, <math>dx=2udu</math> | + | |Hence, <math>dx=2udu.</math> |
|- | |- | ||
|Using all this information, we get | |Using all this information, we get | ||
|- | |- | ||
| − | |<math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math> | + | | <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. | |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 <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 <math>(x-1)(x+1)</math> to get |
|- | |- | ||
| − | |<math>2=A(x-1)+B(x+1).</math> | + | | <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 <math>x=1,</math> we get <math>B=1.</math> |
|- | |- | ||
| − | |If we let <math>x=-1,</math> we get <math>A=-1.</math> | + | |If we let <math>x=-1,</math> we get <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 <math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math> |
|- | |- | ||
|Using this equation, we have | |Using this equation, we have | ||
| Line 155: | Line 155: | ||
!Step 4: | !Step 4: | ||
|- | |- | ||
| − | |To complete this integral, we need to use <math>u</math>-substitution. | + | |To complete this integral, we need to use <math>u</math>-substitution. |
|- | |- | ||
| − | |For the first integral, let <math>t=u+1.</math> Then, <math>dt=du.</math> | + | |For the first integral, let <math>t=u+1.</math> Then, <math>dt=du.</math> |
|- | |- | ||
| − | |For the second integral, let <math>v=u-1.</math> Then, <math>dv=du.</math> | + | |For the second integral, let <math>v=u-1.</math> Then, <math>dv=du.</math> |
|- | |- | ||
|Finally, we integrate to get | |Finally, we integrate to get | ||
Revision as of 10:00, 3 March 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}
| Foundations: |
|---|
| Through partial fraction decomposition, we can write the fraction |
| 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 \frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}} |
| for some constants 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 A,B.} |
Solution:
(a)
| Step 1: |
|---|
| First, we factor the denominator to get |
| 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=\int \frac{3x-1}{x(2x-1)}.} |
| We use the method of partial fraction decomposition. |
| We let |
| 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 \frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.} |
| If we multiply both sides of this equation by 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 x(2x-1),} we get |
| 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 3x-1=A(2x-1)+Bx.} |
| Step 2: |
|---|
| Now, if we let 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 x=0,} we get 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 A=1.} |
| If we let 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 x=\frac{1}{2},} we get 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 B=1.} |
| Therefore, |
| 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 \frac{3x-1}{x(2x-1)}=\frac{1}{x}+\frac{1}{2x-1}.} |
| Step 3: |
|---|
| Therefore, we have |
| 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 \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}} |
| Now, we use 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 u} -substitution. |
| Let 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 u=2x-1.} |
| Then, 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 du=2dx} and 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 \frac{du}{2}=dx.} |
| Hence, we have |
| 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 \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}} |
(b)
| Step 1: |
|---|
| We begin by using 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 u} -substitution. |
| Let 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 u=\sqrt{x+1}.} |
| Then, 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 u^2=x+1} and 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 x=u^2-1.} |
| Also, we have |
| 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 \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}} |
| Hence, 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 dx=2udu.} |
| Using all this information, we get |
| 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=\int \frac{2u^2}{u^2-1}~du.} |
| Step 2: |
|---|
| Now, we have |
| 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 \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}} |
| Step 3: |
|---|
| Now, for the remaining integral, we use partial fraction decomposition. |
| Let 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 \frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.} |
| Then, we multiply this equation by 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 (x-1)(x+1)} to get |
| 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 2=A(x-1)+B(x+1).} |
| If we let 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 x=1,} we get 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 B=1.} |
| If we let 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 x=-1,} we get 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 A=-1.} |
| Thus, we have 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 \frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.} |
| Using this equation, we have |
| 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 \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}} |
| Step 4: |
|---|
| To complete this integral, we need to use 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 u} -substitution. |
| For the first integral, let 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 t=u+1.} Then, 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 dt=du.} |
| For the second integral, let 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 v=u-1.} Then, 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 dv=du.} |
| Finally, we integrate to get |
| 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 \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}} |
| Final Answer: |
|---|
| (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 \ln |x|+\frac{1}{2}\ln |2x-1|+C} |
| (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 2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C} |