Difference between revisions of "009B Sample Final 1, Problem 4"

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<span class="exam"> Compute the following integrals.
 
<span class="exam"> Compute the following integrals.
  
−
<span class="exam">a) <math>\int e^x(x+\sin(e^x))~dx</math>
+
<span class="exam">(a) &nbsp;<math>\int \frac{t^2}{\sqrt{1-t^6}}~dt</math>
  
−
<span class="exam">b) <math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
+
<span class="exam">(b) &nbsp;<math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
−
 
 
−
<span class="exam">c) <math>\int \sin^3x~dx</math>
 
  
 +
<span class="exam">(c) &nbsp;<math>\int \sin^3x~dx</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|Review <math>u</math>-substitution
+
|'''1.''' Through partial fraction decomposition, we can write the fraction
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
 
|-
 
|-
−
|Integration by parts
+
|&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B.</math>
 
|-
 
|-
−
|Partial fraction decomposition
+
|'''2.''' Recall the Pythagorean identity
 
|-
 
|-
−
|Trig identities
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)+\cos^2(x)=1.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|We first distribute to get
+
|We first note that
 
|-
 
|-
 
|
 
|
−
::<math>\int e^x(x+\sin(e^x))~dx=\int e^xx~dx+\int e^x\sin(e^x)~dx</math>.
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{t^2}{\sqrt{1-t^6}}~dt=\int \frac{t^2}{\sqrt{1-(t^3)^2}}~dt.</math>
 
|-
 
|-
−
|Now, for the first integral on the right hand side of the last equation, we use integration by parts.  
+
|Now, we proceed by &nbsp;<math>u</math>-substitution.
 
|-
 
|-
−
|Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=e^x</math>.
+
|Let &nbsp;<math style="vertical-align: 0px">u=t^3.</math> &nbsp;
 +
|-
 +
|Then, &nbsp; <math style="vertical-align: 0px">du=3t^2dt</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{3}=t^2dt.</math>
 
|-
 
|-
 
|So, we have
 
|So, we have
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\int \frac{t^2}{\sqrt{1-t^6}}~dt=\int \frac{1}{3\sqrt{1-u^2}}~du.</math>
−
\displaystyle{\int e^x(x+\sin(e^x))~dx=} & = & \displaystyle{\bigg(xe^x-\int e^x~dx \bigg)+\int e^x\sin(e^x)~dx}\\
 
−
&&\\
 
−
& = & \displaystyle{xe^x-e^x+\int e^x\sin(e^x)~dx}\\
 
−
\end{array}</math>
 
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, for the one remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
+
|Now, we need to use trig substitution.
 
|-
 
|-
−
|Let <math style="vertical-align: 0px">u=e^x</math>. Then, <math style="vertical-align: 0px">du=e^xdx</math>.
+
|Let &nbsp;<math style="vertical-align: -1px">u=\sin \theta.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=\cos \theta d\theta.</math>
 
|-
 
|-
 
|So, we have
 
|So, we have
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
−
\displaystyle{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{xe^x-e^x+\int \sin(u)~du}\\
+
\displaystyle{\int \frac{t^2}{\sqrt{1-t^6}}~dt} & = & \displaystyle{\int \frac{\cos \theta}{3\sqrt{1-\sin^2\theta}}~d\theta}\\
 +
&&\\
 +
& = & \displaystyle{\int \frac{\cos \theta}{3\sqrt{\cos^2\theta}}~d\theta}\\
 +
&&\\
 +
& = & \displaystyle{\int \frac{\cos \theta}{3\cos \theta} d\theta}\\
 +
&&\\
 +
& = & \displaystyle{\int \frac{1}{3}~d\theta}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{3}\theta +C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{xe^x-e^x-\cos(u)+C}\\
+
& = & \displaystyle{\frac{1}{3}\arcsin(u)+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{xe^x-e^x-\cos(e^x)+C}\\
+
& = & \displaystyle{\frac{1}{3}\arcsin(t^3)+C.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we add and subtract <math style="vertical-align: 0px">x</math> from the numerator.  
+
|First, we add and subtract &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; from the numerator.  
 
|-
 
|-
 
|So, we have
 
|So, we have
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& = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 
& = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}\\
+
& = & \displaystyle{\int 1~dx+\int\frac{1-x}{2x^2+x}~dx}.\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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|Now, we need to use partial fraction decomposition for the second integral.
 
|Now, we need to use partial fraction decomposition for the second integral.
 
|-
 
|-
−
|Since <math style="vertical-align: -5px">2x^2+x=x(2x+1)</math>, we let <math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}</math>.
+
|Since &nbsp;<math style="vertical-align: -5px">2x^2+x=x(2x+1),</math>&nbsp; we let  
 
|-
 
|-
−
|Multiplying both sides of the last equation by <math style="vertical-align: -5px">x(2x+1)</math>,
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.</math>
 
|-
 
|-
−
|we get <math style="vertical-align: -5px">1-x=A(2x+1)+Bx</math>.
+
|Multiplying both sides of the last equation by &nbsp;<math style="vertical-align: -5px">x(2x+1),</math>
 
|-
 
|-
−
|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes <math style="vertical-align: -1px">1=A</math>.
+
|we get
 
|-
 
|-
−
|If we let <math style="vertical-align: -14px">x=-\frac{1}{2}</math>, then we get <math style="vertical-align: -14px">\frac{3}{2}=-\frac{1}{2}B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">1-x=A(2x+1)+Bx.</math>
 
|-
 
|-
−
|So, in summation, we have <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
+
|If we let &nbsp;<math style="vertical-align: -5px">x=0,</math> the last equation becomes &nbsp;<math style="vertical-align: -1px">1=A.</math>
 +
|-
 +
|If we let &nbsp;<math style="vertical-align: -14px">x=-\frac{1}{2},</math>&nbsp; then we get &nbsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: 0px">B=-3.</math>
 +
|-
 +
|So, in summation, we have  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
−
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
+
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int 1~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
+
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
+
|For the final remaining integral, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -2px">u=2x+1.</math>&nbsp;
 
|-
 
|-
−
|Let <math style="vertical-align: -2px">u=2x+1</math>. Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
+
|Then, &nbsp;<math style="vertical-align: 0px">du=2\,dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
 
|-
 
|-
−
|Thus, our final integral becomes
+
|Thus, our integral becomes
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 
& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}\\
+
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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|-
 
|-
 
|
 
|
−
::<math>\int \frac{2x^2+1}{2x^2+x}~dx=x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x-\frac{3}{2}\ln (2x+1) +C.}
 +
\end{array}</math>
 
|}
 
|}
 +
 
'''(c)'''
 
'''(c)'''
  
Line 143: Line 162:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
+
|First, we write  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx.</math>
 +
|-
 +
|Using the identity &nbsp;<math style="vertical-align: -5px">\sin^2x+\cos^2x=1,</math>&nbsp; we get
 
|-
 
|-
−
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>  
 
|-
 
|-
 
|If we use this identity, we have
 
|If we use this identity, we have
 
|-
 
|-
−
| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx.</math>
 
|-
 
|-
 
|
 
|
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x</math>. Then, <math style="vertical-align: -1px">du=-\sin x dx</math>.
+
|Now, we proceed by &nbsp;<math>u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math>u=\cos x.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=-\sin x dx.</math>  
 
|-
 
|-
 
|So we have
 
|So we have
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\
 
\displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{-u+\frac{u^3}{3}+C}\\
 
& = & \displaystyle{-u+\frac{u^3}{3}+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}\\
+
& = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}.\\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
 
|
 
|
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|'''(a)''' <math>xe^x-e^x-\cos(e^x)+C</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>\frac{1}{3}\arcsin(t^3)+C</math>
 
|-
 
|-
−
|'''(b)''' <math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
 
|-
 
|-
−
|'''(c)''' <math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
+
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;<math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 13:01, 20 May 2017

Compute 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{t^2}{\sqrt{1-t^6}}~dt}

(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{2x^2+1}{2x^2+x}~dx}

(c)  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 \sin^3x~dx}

Foundations:  
1. 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.}
2. Recall the Pythagorean identity
       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 \sin^2(x)+\cos^2(x)=1.}


Solution:

(a)

Step 1:  
We first note that

       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{t^2}{\sqrt{1-t^6}}~dt=\int \frac{t^2}{\sqrt{1-(t^3)^2}}~dt.}

Now, we proceed 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 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=t^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=3t^2dt}   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}{3}=t^2dt.}
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{t^2}{\sqrt{1-t^6}}~dt=\int \frac{1}{3\sqrt{1-u^2}}~du.}

Step 2:  
Now, we need to use trig 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=\sin \theta.}   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=\cos \theta d\theta.}
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{t^2}{\sqrt{1-t^6}}~dt} & = & \displaystyle{\int \frac{\cos \theta}{3\sqrt{1-\sin^2\theta}}~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{\cos \theta}{3\sqrt{\cos^2\theta}}~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{\cos \theta}{3\cos \theta} d\theta}\\ &&\\ & = & \displaystyle{\int \frac{1}{3}~d\theta}\\ &&\\ & = & \displaystyle{\frac{1}{3}\theta +C}\\ &&\\ & = & \displaystyle{\frac{1}{3}\arcsin(u)+C}\\ &&\\ & = & \displaystyle{\frac{1}{3}\arcsin(t^3)+C.} \end{array}}

(b)

Step 1:  
First, we add and subtract  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}   from the numerator.
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{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int\frac{2x^2+x-x+1}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int 1~dx+\int\frac{1-x}{2x^2+x}~dx}.\\ \end{array}}
Step 2:  
Now, we need to use partial fraction decomposition for the second integral.
Since  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 2x^2+x=x(2x+1),}   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{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.}
Multiplying both sides of the last 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 1-x=A(2x+1)+Bx.}
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,} the last equation becomes  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 1=A.}
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},}   then 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 \frac{3}{2}=-\frac{1}{2}\,B.}   Thus,  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=-3.}
So, in summation, 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{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.}
Step 3:  
If we plug in the last equation from Step 2 into our final integral in Step 1, 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{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int 1~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx.}\\ \end{array}}

Step 4:  
For the final remaining integral, 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=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.}
Thus, our integral becomes

       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{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\ &&\\ & = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C.}\\ \end{array}}

Therefore, the final answer is

       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{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x-\frac{3}{2}\ln (2x+1) +C.} \end{array}}

(c)

Step 1:  
First, we write
       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\sin^3x~dx=\int \sin^2 x \sin x~dx.}
Using the identity  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 \sin^2x+\cos^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 \sin^2x=1-\cos^2x.}
If we use this identity, 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\sin^3x~dx=\int (1-\cos^2x)\sin x~dx.}
Step 2:  
Now, we proceed 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 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=\cos 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=-\sin x dx.}
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\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\ &&\\ & = & \displaystyle{-u+\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+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 \frac{1}{3}\arcsin(t^3)+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 x+\ln x-\frac{3}{2}\ln (2x+1) +C}
    (c)    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 x+\frac{\cos^3x}{3}+C}

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