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\begin{document}

   \begin{center}
     {\bf MTH112 -- TEST 3 Solutions}
   \end{center}

\begin{flushright}
   Name: \rule[0in]{1.5in}{.01in}
\end{flushright} \vs

\noin HONOR CODE: On my honor, I have neither given nor received 
any aid on this examination. \\
Signature: \rule[0in]{1.5in}{.01in}

\newcommand{\onethird}{\frac{1}{3}}

\noin Note: Show all work on exam in order to receive full credit. 

\begin{enumerate}
\item  Integrate:
	\begin{enumerate}
	
	\item $\ds \int_3^7 \frac{1}{(x+1)(x-2)} \; dx $
	\begin{eqnarray*}
	  A/(x+1) + B/(x-2) &=& 1/(x+1)(x-2) \\
	  A(x-2) + B(x+1) &=& 1 \\
	  x=2 &\Rightarrow& B = 1/3 \\
	  x=-1 &\Rightarrow& A = -1/3 \\
	  \end{eqnarray*}
	  Hence,
	  \begin{eqnarray*}
	  \int_3^7 \frac{1}{(x+1)(x-2)} \; dx  &=& (-1/3)(\ln|x+1| - \ln|x-2|) \big]^7_3\\
	   &= & \onethird \left.\ln \left|\frac{ x-2}{x+1} \right| \right]_3^7 \\
	   &=& \onethird ( \ln 5/8 - \ln 1/4 ) = \frac{ \ln (5/2)}{3}.
	\end{eqnarray*}
	
	\item $\ds  \int \frac{1}{x(x+1)(x+2)} \; dx $
	\begin{eqnarray*}
	 A/x + B/(x+1) + C/(x+2) &=& 1/x(x+1)(x+2) \\
	 A(x+1)(x+2) + Bx(x+2) + Cx(x+1) &=& 1 \\
	 x=-1 &\Rightarrow& B = -1 \\
	 x=-2  &\Rightarrow&  C = 1/2 \\
	 x=0  &\Rightarrow& A = 1/2 \\
	 \end{eqnarray*}
	  Hence,
	  \begin{eqnarray*}
	  \int \frac{1}{x(x+1)(x+2)} \; dx &=& (1/2)(\ln|x| + \ln|x+2| )- \ln|x+1| + C \\
	 %  &=& \ln |\sqrt{x(x+2)}/(x+1)| + C.
	\end{eqnarray*}
	
	\item $\ds \int \frac {x}{(x-3)(x+2)^2} \; dx $
	\begin{eqnarray*}
	B/(x-3)+C/(x+2)+D/(x+2)^2 &=& x/((x-3)(x+2)^2) \\
	B(x+2)^2   C(x-3)(x+2) + D(x-3) &=& x \\
	x=3 &\Rightarrow& B = 3/25 \\
	x = -2 &\Rightarrow& D = 2/5 \\
	0x^2 &\Rightarrow& C = -B \\
	\end{eqnarray*}
	Hence,
	\[ \int \frac {x}{(x-3)(x+2)^2} \; dx = 
	\frac{3}{25}\ln\left|\frac{x-3}{x+2}\right| + \frac{-2}{5(x+2)} + C \]
	
	\item $\ds \int \frac{5x^2+3x-2}{x^3+2x^2} \; dx $
	\begin{eqnarray*}
	A/x + Bx^2 + C/(x+2) &=& (5x^2+3x-2)/(x^3+2x^2) \\
	Ax(x+2) + B(x+2) + Cx^2 &=& 5x^2+3x-2 \\
	x = 0 &\Rightarrow& B = -1 \\
	x = -2 &\Rightarrow& C = 3 \\
	5x^2 &\Rightarrow&  A + C = 5  \Rightarrow A = 3\\
	\end{eqnarray*}
	Hence,
	\begin{eqnarray*}
	\int \frac{5x^2+3x-2}{x^3+2x^2} \; dx &=&  \int 2/x \,dx + 
		\int -1/x^2\, dx + 3 \int 1/(x+2)\,dx \\
		&=& 2\ln|x| + (1/x) + 3\ln|x+2| + C
	\end{eqnarray*}	
	\end{enumerate}

\newpage
\item Integrate:
	\begin{enumerate}
	
	\item $\ds \int x \sqrt{4-x^2} \; dx $;
	Let $u=4-x^2$, then $du = -2xdx$, so,
	\begin{eqnarray*}
	\int x \sqrt{4-x^2} \; dx  &=& (-1/2)\int \sqrt{u}du 
	\\ &=& (-1/3) u^{3/2} + C
	\\ &=& \frac{-1(4-x^2)^{3/2}}{3} + C \\
	\end{eqnarray*}	
	
	\item $\ds \int \sqrt{3-2x-x^2} \; dx $;
	Complete the square:
	\[ 3-2x-x^2 = -(x+1)^2 + 4 \]
	Let $u=x+1$, $du=dx$, then use formula 30:
	\begin{eqnarray*}
	\int \sqrt{3-2x-x^2} \; dx &=& \int \sqrt{4-u^2}du \\
	&=&  (u/2)\sqrt{4-u^2} + (4/2)\sin^{-1}(u/2)+ C \\
	&=& \frac{(x+1) \sqrt{3-2x-x^2}}{2} + 2\sin^{-1}\frac{x+1}{2} + C
	\end{eqnarray*}
	
	\item $\ds \int \frac{\sqrt{9x^2-4}}{x}\; dx $; Let $3x=u$ then use
	formula 41:
	\begin{eqnarray*}
	\int \bigl(\sqrt{9x^2-4}\bigr)/x\; dx &=& \int \bigl(\sqrt{u^2-4}\bigr)/u\;du \\
	&=& \sqrt{u^2-4} - 2\cos^{-1}(2/u) + C \\
	&=& \sqrt{9x^2-4} - 2 \tan^{-1}\frac{\sqrt{9x^2-4}}{2} + C
	\end{eqnarray*}
	
	\item $\ds \int \frac{x}{(x^2+4)^{5/2}} \; dx$; Let $u=x^2+4, du = 2xdx$:
	\begin{eqnarray*}
	(1/2)\int 1/u^{5/2}\, du &=& (-1/3)\left( 1/u^{3/2} \right) + C \\
	&=& \frac{-1}{3(x^2+4)^{3/2}} + C
	\end{eqnarray*}
	\end{enumerate}

\newpage
\item Integrate:
	\begin{enumerate}
	
	\item $\ds \int \sin x + x^5  \; dx $
	\begin{eqnarray*}
	\int \sin x + x^5  \; dx  &=& \int \sin x \, dx + \int x^5 \, dx \\
	&=& -\cos x + (1/6)x^6 + C 
	\end{eqnarray*}
	
	\item $\ds \int \frac{1+\sqrt{x+4}}{x} \; dx $
	\begin{eqnarray*}
	 \int \frac{1+\sqrt{x+4}}{x} \; dx &=& \int 1/x \, dx + \int \sqrt{x+4}/x \, dx \\
	 &=& 
	 \int 1/x \, dx  + \int \frac{x+4}{x\sqrt{x+4}}\,dx \\
	 &=& \int 1/x \, dx  + \int \frac{1}{\sqrt{x+4}}\, dx + 4 \int \frac{1}{x\sqrt{x+4}}\,dx
	\end{eqnarray*}
	Integrate term by term, using formula 57 for the last integral:
	\[ \int \frac{1+\sqrt{x+4}}{x} \, dx  = \ln|x| + 2 \sqrt{x+4} + 2 \ln\left|
	\frac{\sqrt{x+4}-2}{\sqrt{x+4+2}} \right| + C
	\]
	
	\item $\ds \int \pi \sin x + \cos(1/2) \tan(\pi x) \; dx $
	\begin{eqnarray*}
	\int \pi \sin x + \cos(1/2) \tan(\pi x) \, dx &=&
	   \pi \int \sin x \, dx + \cos(1/2) \int \tan u \, (du/\pi) \\
	&=& -\pi \cos x + \cos(1/2)\ln |\sec u|/\pi  + C \\
	&=& -\pi \cos x +  \cos(1/2) \ln |\sec (\pi x)|/\pi  + C
	\end{eqnarray*}
	
 	\item $\ds \int (1+\sqrt{x})^2 \; dx$
	\begin{eqnarray*}
	\int (1+\sqrt{x})^2 \, dx &=& \int 1 + 2\sqrt{x} + x \, dx \\
	&=& x + (4/3)x^{3/2}+x^2/2 + C 
	\end{eqnarray*}
        \end{enumerate}

\newpage
\item  Integrate:
	\begin{enumerate}
	
    	\item $\ds \int \frac{1}{x-\root 3 \of {x}}\;dx$; Use a rationalizing
	substituion $u = \root 3 \of {x}, u^3 = x, 3u^2 du = dx$:, 
	\begin{eqnarray*}
	\int \frac{3u^2}{u^3 - u}\, du &=& 3 \int u/(u^2-1) \, du \\
	&=& 3 \int A/(u-1) + B/(u+1)\, du \\
	&=& (3/2) \int 1/(u-1) + 1/(u+1) \,du \\
	&=& (3/2) \ln |u^2-1| + C \\
	&=& \frac{3\ln |x^{2/3}-1|}{2} + C \\
	\end{eqnarray*}
	
	\item $\ds \int_1^3 \frac{\sqrt{x-1}}{x+1} \; dx$; Let $u=\sqrt{x-1},
	u^2+1 =x, 2u\,du = dx$, then substitute $\sqrt{2}w=u$:
	\begin{eqnarray*}
	 \int_1^3 \frac{\sqrt{x-1}}{x+1} \, dx
	  &=& \int_0^{\sqrt{2}} \frac{u}{u^2+2} 2u\, du \\
	  &=& 2\sqrt{2}\int_0^1 \frac{w^2}{w^2+1} \, dw \\
	  &=& 2\sqrt{2}\int_0^1 1 - 1/(w^2+1) \, dw \\
	  &=& 2\sqrt{2} \left( w - \tan^{-1} w  \right)^1_0 \\
	  &=& 2\sqrt{2} ( 1 - \pi/4 )
	\end{eqnarray*}
	
	\item $\ds \int \frac{1}{\sqrt{1+\sqrt{x}}} \; dx$;
	Let $u=\sqrt{x}, u^2=x, 2udu=dx$, then use formula 55:
	\begin{eqnarray*}
	\int \frac{1}{\sqrt{1+\sqrt{x}}} \, dx &=& 2\int \frac{u\,du}{\sqrt{1+u}} \\
	&=& 2( (2/3)(u-2)\sqrt{1+u}) + C \\
	&=& (4/3)(\sqrt{x}-2)\sqrt{1+\sqrt{x}} + C
	\end{eqnarray*}
\newpage
	\item $\ds \int \frac{\root 3 \of {x} +1}
			{\root 3 \of {x} -1} \; dx$; Use a rationalizing substitution,
	divide and integrate term by term:
	\begin{eqnarray*}
	\int \frac{\root 3 \of {x} +1}{\root 3 \of {x} -1} \, dx
	&=& \int \frac{u+1}{u-1} 3u^2\, du \\
	&=& 3\int u^2 + 2u + 2 + 2/(u-1) \, du \\
	&=& u^3 + 3u^2 + 6u + 6\ln|u-1| + C \\
	&=& x + 3 (\root 3 \of {x})^2 + 6\root 3 \of {x} + 6\ln|\root 3 \of {x} -1| +C
	\end{eqnarray*}
 	\end{enumerate}

%\newpage
\item
	\begin{enumerate}
	
	\item Factor $x^4+1$ as the difference of squares by first adding
	add subtracting the same quantity. Use this factorization to
	evaluate: \[\int \frac{1}{x^4+1}\;dx\]
	The factorization is 
	\[x^4+1 = (x^2+1)^2 -2x^2 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x + 1). \]
	Use partial fractions, complete the squares, and integrate term by term.
	I used the computer program {\em Mathematica} to calculate the solution:
	\begin{eqnarray*}
	\int \frac{1}{x^4+1}\,dx &=& 
	(   
	  2\tan^{-1}(\sqrt{2}x-1) + 2\tan^{-1}(\sqrt{2}x+1) \\
	  & &- \ln | 1-\sqrt{2}x+x^2| + \ln|1+\sqrt{2}x+x^2|
	)/(4\sqrt{2}) +C
	\end{eqnarray*}
	
	\item Integrate: \[\int e^x \cos(3x+4) \;dx\]
	Substitute $u=3x+4$, factor out the $e^{k}$ and use formula 99:
	\begin{eqnarray*}
	\int e^x \cos(3x+4) \,dx &=& \int e^{(u-4)/3} \cos u \, (du/3) \\
	&=& (1/3)e^{-4/3}\int e^{u/3}\cos u \, du \\\
	&=& (1/3)e^{-4/3}\left( \frac{e^{u/3}} {(1/3)^2+1} ((1/3)\cos u + \sin u)\right) + C 
	\\ &=& (3e^x/10)\left( (1/3)\cos(3x+4) + \sin(3x+4)\right) + C
	\end{eqnarray*}
	
\newpage
	\item Write out the form of the partial fraction
	decomposition for this function: --- do {\em not} determine the numerical values
	of the cofficients: \[\frac{1}{x^6-x^3}\]
	\begin{eqnarray*}
	\frac{1}{x^6-x^3} &=& \frac{1}{x^3(x-1)(x^2+x+1)} \\
	&=& A/x + B/x^2 + C/x^3 + D/(x-1) + (Ex+F)/(x^2+x+1)
	\end{eqnarray*}
	
	\item Find the limit:\[\lim_{x\rightarrow \infty}\;
		\frac{\ln\ln x }{ \sqrt{x} }\]
	Use L'Hopital's:
	\begin{eqnarray*}
	\lim_{x\rightarrow \infty}\;\frac{\ln\ln x }{ \sqrt{x} } 
	&=& \lim_{x\rightarrow \infty}\;\frac{(1/\ln x)(1/x)}{1/(-2\sqrt{x})} \\
	&=& \lim_{x\rightarrow \infty}\;\frac{-2}{\sqrt{x}\ln x} = 0 
	\end{eqnarray*}	
	\end{enumerate}
 
\end{enumerate}
\end{document}




