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==
Line 14: Line 20:
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let <math>u=4x^2+5x+3</math>. Then, <math>du=(8x+5)~dx</math>.
+
|Let &nbsp;<math style="vertical-align: -3px">u=4x^2+5x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">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 style="vertical-align: -14px">\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;
 
|-
 
|-
−
|<math>e^{4x^2+5x+3}+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>e^{4x^2+5x+3}+C</math>
 
|-
 
|-
 
|}
 
|}
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -2px">u=1-2x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -2px">du=-4x~dx.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: -15px">\frac{du}{-4}=x~dx.</math>&nbsp;
 +
|-
 +
|Plugging these into our integral, we get
 
|-
 
|-
 
|
 
|
 +
::<math>\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}</math>
 
|-
 
|-
 
|}
 
|}
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!Final Answer: &nbsp;
 
!Final Answer: &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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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -6px">u=\ln(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -14px">du=\frac{1}{x}~dx.</math>
 +
|-
 +
|Plugging these into our integral, we get
 
|-
 
|-
 
|
 
|
 +
::<math>\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}</math>
 
|-
 
|-
 
|}
 
|}
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|<math>-\cos(\ln x)+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\cos(\ln x)+C</math>
 
|-
 
|-
 
|}
 
|}
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -1px">u=x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2x~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">\frac{du}{2}=x~dx.</math>&nbsp;
 +
|-
 +
|Plugging these into our integral, we get
 
|-
 
|-
 
|
 
|
 +
::<math>\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}</math>
 
|-
 
|-
 
|}
 
|}
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!Final Answer: &nbsp;
 
!Final Answer: &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 &nbsp;<math style="vertical-align: -16px">\int \frac{2}{y^2+4}~dy.</math>
 +
 +
First, we factor out &nbsp;<math style="vertical-align: -1px">4</math>&nbsp; out of the denominator.
 +
 +
So, we have
 +
 +
::<math>\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}</math>
 +
 +
Now, we use &nbsp;<math style="vertical-align: -1px">u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
 +
 +
Then, &nbsp;<math style="vertical-align: -14px">du=\frac{1}{2}~dy</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">2~du=dy.</math>
 +
 +
Plugging these into our integral, we get
 +
 +
::<math>\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}</math>
  
 +
So, we have
 +
::<math>\int \frac{2}{y^2+4}~dy=\arctan\bigg(\frac{y}{2}\bigg)+C.</math>
  
 
== Exercise 2 ==
 
== Exercise 2 ==
  
 +
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -17px">\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>
  
 +
Let &nbsp;<math style="vertical-align: -5px">u=5+\sin(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">u=\cos(x)~dx.</math>
  
 +
Plugging these into our integral, we get
 +
 +
::<math>\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}</math>
 +
 +
So, we have
 +
::<math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx=-\frac{1}{5+\sin(x)}+C.</math>
  
 
== Exercise 3 ==
 
== Exercise 3 ==
  
 +
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -16px">\int \frac{x+5}{2x+3}~dx.</math>
 +
 +
Here, the substitution is not obvious.
 +
 +
Let &nbsp;<math style="vertical-align: -3px">u=2x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
 +
 +
Now, we need a way of getting rid of &nbsp;<math style="vertical-align: -2px">x+5</math>&nbsp; in the numerator.
  
 +
Solving for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in the first equation, we get &nbsp;<math style="vertical-align: -14px">x=\frac{1}{2}u-\frac{3}{2}.</math>
 +
 +
Plugging these into our integral, we get
 +
 +
::<math>\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}</math>
 +
 +
So, we get
 +
::<math>\int \frac{x+5}{2x+3}~dx=\frac{1}{4}(2x+3+7\ln|2x+3|)+C.</math>
  
 
== Exercise 4 ==
 
== Exercise 4 ==
 +
 +
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -14px">\int \frac{x^2+4}{x+2}~dx.</math>
 +
 +
Let &nbsp;<math style="vertical-align: -2px">u=x+2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=dx.</math>
 +
 +
Now, we need a way of replacing &nbsp;<math style="vertical-align: -2px">x^2+4.</math>
 +
 +
If we solve for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in our first equation, we get &nbsp;<math style="vertical-align: -1px">x=u-2.</math>
 +
 +
Now, we square both sides of this last equation to get &nbsp;<math style="vertical-align: -5px">x^2=(u-2)^2.</math>
 +
 +
Plugging in to our integral, we get
 +
 +
::<math>\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}</math>
 +
 +
So, we have
 +
::<math>\int \frac{x^2+4}{x+2}~dx=\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.</math>

Latest revision as of 14:57, 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  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 method and

(2) correctly identify the necessary substitution.

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

Warm-Up

Evaluate the following indefinite integrals.

1)   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 (8x+5)e^{4x^2+5x+3}~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=4x^2+5x+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=(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
Final Answer:  
       

2)  

Solution:  
Let    Then,    Hence,   
Plugging these into our integral, we get
Final Answer:  
       

3)  

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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