Difference between revisions of "U-substitution"

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==Introduction==
 
==Introduction==
−
The method of <math>u</math>-substitution is used to simplify the function you are integrating so that you can easily recognize it's antiderivative. This method is closely related to the chain rule for derivatives.
+
The method of &nbsp;<math style="vertical-align: -1px">u</math>-substitution is used to simplify the function you are integrating so that you can easily recognize it's antiderivative.  
  
−
One question that is frequently asked is "How do you know what substitution to make?" In general, this is a difficult question to answer since it is dependent on the integral. The best way to master <math>u</math>-substitution is to work out as many problems as possible. This will help you: (1) understand the <math>u</math>-substitution method and (2) correctly identify the necessary substitution.
+
This method is closely related to the chain rule for derivatives.
  
−
<u>NOTE</u>: After you plug-in your substitution, all of the <math>x</math>'s in your integral should be gone. The only variables remaining in your integral should be <math>u</math>'s.
+
One question that is frequently asked is "How do you know what substitution to make?" In general, this is a difficult question to answer since it is dependent on the integral. The best way to master &nbsp;<math style="vertical-align: -1px">u</math>-substitution is to work out as many problems as possible. This will help you:
 +
 
 +
(1) understand the &nbsp;<math style="vertical-align: -1px">u</math>-substitution method and
 +
 
 +
(2) correctly identify the necessary substitution.
 +
 
 +
<u>NOTE</u>: After you plug-in your substitution, all of the &nbsp;<math style="vertical-align: 0px">x</math>'s in your integral should be gone. The only variables remaining in your integral should be &nbsp;<math style="vertical-align: 0px">u</math>'s.
  
 
==Warm-Up==
 
==Warm-Up==
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let <math>u=4x^2+5x+3</math>. Then, <math>du=(8x+5)~dx</math>.
+
|Let &nbsp;<math>u=4x^2+5x+3.</math>&nbsp; Then, &nbsp;<math>du=(8x+5)~dx</math>.
 
|-
 
|-
 
|-
 
|-
−
|Plugging these into our integral, we get <math>\int e^u~du</math>, which we know how to integrate.
+
|Plugging these into our integral, we get &nbsp;<math>\int e^u~du,</math>&nbsp; which we know how to integrate.
 
|-
 
|-
 
|So, we get
 
|So, we get
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& = & \displaystyle{e^u+C}\\
 
& = & \displaystyle{e^u+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{e^{4x^2+5x+3}+C}. \\
+
& = & \displaystyle{e^{4x^2+5x+3}+C.} \\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp;<math>e^{4x^2+5x+3}+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>e^{4x^2+5x+3}+C</math>
 
|-
 
|-
 
|}
 
|}
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let <math>u=1-2x^2</math>. Then, <math>du=-4x~dx</math>. Hence, <math>\frac{du}{-4}=x~dx</math>.
+
|Let &nbsp;<math>u=1-2x^2.</math>&nbsp; Then, &nbsp;<math>du=-4x~dx.</math>&nbsp; Hence, &nbsp;<math>\frac{du}{-4}=x~dx.</math>&nbsp;
 
|-
 
|-
 
|Plugging these into our integral, we get
 
|Plugging these into our integral, we get
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& = & \displaystyle{-\frac{1}{2}u^{\frac{1}{2}}+C}\\
 
& = & \displaystyle{-\frac{1}{2}u^{\frac{1}{2}}+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\frac{1}{2}\sqrt{1-2x^2}+C}. \\
+
& = & \displaystyle{-\frac{1}{2}\sqrt{1-2x^2}+C.} \\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp;<math>-\frac{1}{2}\sqrt{1-2x^2}+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\frac{1}{2}\sqrt{1-2x^2}+C</math>
 
|-
 
|-
 
|}
 
|}
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let <math>u=\ln(x)</math>. Then, <math>du=\frac{1}{x}~dx</math>.
+
|Let &nbsp;<math>u=\ln(x).</math>&nbsp; Then, &nbsp;<math>du=\frac{1}{x}~dx.</math>
 
|-
 
|-
 
|Plugging these into our integral, we get
 
|Plugging these into our integral, we get
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& = & \displaystyle{-\cos(u)+C}\\
 
& = & \displaystyle{-\cos(u)+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\cos(\ln x)+C}. \\
+
& = & \displaystyle{-\cos(\ln x)+C.} \\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp;<math>-\cos(\ln x)+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\cos(\ln x)+C</math>
 
|-
 
|-
 
|}
 
|}
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let <math>u=x^2</math>. Then, <math>du=2x~dx</math> and <math>\frac{du}{2}=x~dx</math>.
+
|Let &nbsp;<math>u=x^2.</math>&nbsp; Then, &nbsp;<math>du=2x~dx</math>&nbsp; and &nbsp;<math>\frac{du}{2}=x~dx.</math>&nbsp;
 
|-
 
|-
 
|Plugging these into our integral, we get
 
|Plugging these into our integral, we get
Line 107: Line 113:
 
& = & \displaystyle{\frac{1}{2}e^u+C}\\
 
& = & \displaystyle{\frac{1}{2}e^u+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{1}{2}e^{x^2}+C}. \\
+
& = & \displaystyle{\frac{1}{2}e^{x^2}+C.} \\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp;<math>\frac{1}{2}e^{x^2}+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{2}e^{x^2}+C</math>
 
|-
 
|-
 
|}
 
|}
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== Exercise 1 ==
 
== Exercise 1 ==
  
−
Evaluate the indefinite integral <math>\int \frac{2}{y^2+4}~dy</math>.
+
Evaluate the indefinite integral &nbsp;<math>\int \frac{2}{y^2+4}~dy.</math>
  
−
First, we factor out <math>4</math> out of the denominator.
+
First, we factor out &nbsp;<math>4</math>&nbsp; out of the denominator.
  
 
So, we have
 
So, we have
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\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{4}\int \frac{2}{\frac{y^2}{4}+1}~dy}\\
 
\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{4}\int \frac{2}{\frac{y^2}{4}+1}~dy}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy}.\\
+
& = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy.}\\
 
\end{array}</math>
 
\end{array}</math>
  
−
Now, we use <math>u</math>-substitution. Let <math>u=\frac{y}{2}</math>.
+
Now, we use &nbsp;<math>u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
  
−
Then, <math>du=\frac{1}{2}~dy</math> and <math>2~du=dy</math>.
+
Then, &nbsp;<math>du=\frac{1}{2}~dy</math>&nbsp; and &nbsp;<math>2~du=dy.</math>
  
 
Plugging these into our integral, we get
 
Plugging these into our integral, we get
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& = & \displaystyle{\arctan(u)+C}\\
 
& = & \displaystyle{\arctan(u)+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C}.\\
+
& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C.}\\
 
\end{array}</math>
 
\end{array}</math>
  

Revision as of 12:19, 24 August 2017

Introduction

The method of  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 is used to simplify the function you are integrating so that you can easily recognize it's antiderivative.

This method is closely related to the chain rule for derivatives.

One question that is frequently asked is "How do you know what substitution to make?" In general, this is a difficult question to answer since it is dependent on the integral. The best way to master  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 is to work out as many problems as possible. This will help you:

(1) understand the  -substitution method and

(2) correctly identify the necessary substitution.

NOTE: After you plug-in your substitution, all of the  's in your integral should be gone. The only variables remaining in your integral should be  's.

Warm-Up

Evaluate the following indefinite integrals.

1)  

Solution:  
Let    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=(8x+5)~dx} .
Plugging these into our integral, 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 e^u~du,}   which we know how to integrate.
So, 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 \begin{array}{rcl} \displaystyle{\int (8x+5)e^{4x^2+5x+3}~dx} & = & \displaystyle{\int e^u~du}\\ &&\\ & = & \displaystyle{e^u+C}\\ &&\\ & = & \displaystyle{e^{4x^2+5x+3}+C.} \\ \end{array}}
Final Answer:  
       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 e^{4x^2+5x+3}+C}

2)   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{x}{\sqrt{1-2x^2}}~dx}

Solution:  
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=1-2x^2.}   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=-4x~dx.}   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 \frac{du}{-4}=x~dx.}  
Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int\frac{x}{\sqrt{1-2x^2}}~dx} & = & \displaystyle{\int -\frac{1}{4}~u^{-\frac{1}{2}}~du}\\ &&\\ & = & \displaystyle{-\frac{1}{2}u^{\frac{1}{2}}+C}\\ &&\\ & = & \displaystyle{-\frac{1}{2}\sqrt{1-2x^2}+C.} \\ \end{array}}
Final Answer:  
       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}{2}\sqrt{1-2x^2}+C}

3)   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{\sin(\ln x)}{x}~dx}

Solution:  
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=\ln(x).}   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=\frac{1}{x}~dx.}
Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int\frac{\sin(\ln x)}{x}~dx} & = & \displaystyle{\int \sin(u)~du}\\ &&\\ & = & \displaystyle{-\cos(u)+C}\\ &&\\ & = & \displaystyle{-\cos(\ln x)+C.} \\ \end{array}}
Final Answer:  
       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 -\cos(\ln x)+C}

4)   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 xe^{x^2}~dx}

Solution:  
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=x^2.}   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=2x~dx}   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}=x~dx.}  
Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int xe^{x^2}~dx} & = & \displaystyle{\int \frac{1}{2}e^u~du}\\ &&\\ & = & \displaystyle{\frac{1}{2}e^u+C}\\ &&\\ & = & \displaystyle{\frac{1}{2}e^{x^2}+C.} \\ \end{array}}
Final Answer:  
       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}{2}e^{x^2}+C}

Exercise 1

Evaluate the indefinite integral  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{2}{y^2+4}~dy.}

First, we factor out  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 4}   out of the denominator.

So, 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{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{4}\int \frac{2}{\frac{y^2}{4}+1}~dy}\\ &&\\ & = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy.}\\ \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=\frac{y}{2}.}

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=\frac{1}{2}~dy}   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 2~du=dy.}

Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{2}\int \frac{2}{u^2+1}~du}\\ &&\\ & = & \displaystyle{\int \frac{1}{u^2+1}~du}\\ &&\\ & = & \displaystyle{\arctan(u)+C}\\ &&\\ & = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C.}\\ \end{array}}

So, 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 \int \frac{2}{y^2+4}~dy=\arctan\bigg(\frac{y}{2}\bigg)+C.}

Exercise 2

Evaluate the indefinite integral 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{\cos(x)}{(5+\sin x)^2}~dx.}

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=5+\sin(x).} 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=\cos(x)~dx.}

Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int \frac{\cos(x)}{(5+\sin x)^2}~dx} & = & \displaystyle{\int \frac{1}{u^2}~du}\\ &&\\ & = & \displaystyle{-\frac{1}{u}+C}\\ &&\\ & = & \displaystyle{-\frac{1}{5+\sin(x)}+C.} \end{array}}

So, 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 \int \frac{\cos(x)}{(5+\sin x)^2}~dx=-\frac{1}{5+\sin(x)}+C.}

Exercise 3

Evaluate the indefinite integral 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{x+5}{2x+3}~dx.}

Here, the substitution is not obvious.

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+3} . 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=2~dx} 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} .

Now, we need a way of getting rid of 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+5} in the numerator.

Solving for 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} in the first equation, 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 x=\frac{1}{2}u-\frac{3}{2}} .

Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int \frac{x+5}{2x+3}~dx} & = & \displaystyle{\int \frac{(\frac{1}{2}u-\frac{3}{2})+5}{2u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{2}\int \frac{\frac{1}{2}u+\frac{7}{2}}{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}\int \frac{u+7}{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}\int 1+\frac{7}{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}(u+7\ln|u|)+C}\\ &&\\ & = & \displaystyle{\frac{1}{4}(2x+3+7\ln|2x+3|)+C.}\\ \end{array}}

So, 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{x+5}{2x+3}~dx=\frac{1}{4}(2x+3+7\ln|2x+3|)+C.}

Exercise 4

Evaluate the indefinite integral 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{x^2+4}{x+2}~dx.}

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=x+2} . 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=dx} .

Now, we need a way of replacing 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^2+4} .

If we solve for 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} in our first equation, 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 x=u-2.}

Now, we square both sides of this last equation 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 x^2=(u-2)^2.}

Plugging in to our integral, 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 \begin{array}{rcl} \displaystyle{\int \frac{x^2+4}{x+2}~dx} & = & \displaystyle{\int \frac{(u-2)^2+4}{u}~du}\\ &&\\ & = & \displaystyle{\int \frac{u^2-4u+4+4}{u}~du}\\ &&\\ & = & \displaystyle{\int \frac{u^2-4u+8}{u}~du}\\ &&\\ & = & \displaystyle{\int u-4+\frac{8}{u}~du}\\ &&\\ & = & \displaystyle{\frac{u^2}{2}-4u+8\ln|u|+C}\\ &&\\ & = & \displaystyle{\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.}\\ \end{array}}

So, 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 \int \frac{x^2+4}{x+2}~dx=\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.}
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