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

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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>
 
|&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>
 
|}
 
|}
  
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|&nbsp; &nbsp; &nbsp; &nbsp;<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>
 
|-
 
|-
−
|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>
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>3x-1=A(2x-1)+Bx.</math>
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!Step 2: &nbsp;
 
!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,  
 
|Therefore,  
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\end{array}</math>
 
\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
 
|Hence, we have
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!Step 1: &nbsp;  
 
!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  
 
|Also, we have  
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\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
−
|Hence, <math>dx=2udu</math>.
+
|Hence, &nbsp;<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>
+
|&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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|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 &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
 
|Using this equation, we have
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!Step 4: &nbsp;
 
!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
 
|Finally, we integrate to get

Revision as of 10:00, 3 March 2017

Find the following integrals

(a)  

(b)  

Foundations:  
Through partial fraction decomposition, we can write the fraction
       
for some constants Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A,B.}


Solution:

(a)

Step 1:  
First, we factor the denominator to get
       
We use the method of partial fraction decomposition.
We let
       
If we multiply both sides of this equation by  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x(2x-1),}   we get
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 3x-1=A(2x-1)+Bx.}
Step 2:  
Now, if we let    we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A=1.}
If we let    we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle B=1.}
Therefore,
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {3x-1}{x(2x-1)}}={\frac {1}{x}}+{\frac {1}{2x-1}}.}
Step 3:  
Therefore, we have
       
Now, we use  -substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=2x-1.}
Then,    and  
Hence, we have
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  -substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u={\sqrt {x+1}}.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u^{2}=x+1}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=u^{2}-1.}
Also, we have
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}

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