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\noindent \hfill
\fbox{\fbox{\parbox{7.25in}{\scshape{Math 1110 \hfill Integral Practice}}}} \hfill \\

\begin{center}
\begin{tabular}{lcl}
$\displaystyle 1. \quad \int 3x^2-\frac{1}{x}+e^x-7\,dx $
& \qquad \qquad &
$\displaystyle 2. \quad \int \frac{\sqrt{x}}{x^3} + \frac{6+2x}{x^2}-50x^{99}\,dx $
\vspace{0.15in}\\
$\displaystyle 3. \quad \int 3\sin(x)+10\sec(x)\tan(x)\,dx $
& \qquad \qquad &
$\displaystyle 4. \quad \int \dfrac{5}{\sqrt{1-x^2}}\,dx $
\vspace{0.15in}\\
$\displaystyle 5. \quad \int \,dx $ 
& \qquad \qquad &
$\displaystyle 6. \quad \int \frac{e^{5x}}{e^{2x}} - 8x^{7/3}\,dx $
\vspace{0.15in}\\
$\displaystyle 7. \quad \int (x^2-5x+6)^{9}(2x-5)\,dx $
& \qquad \qquad &
$\displaystyle 8. \quad \int e^{6x-2}\,dx $
\vspace{0.15in}\\
$\displaystyle 9. \quad \int \cos(5x+1)\,dx $
& \qquad \qquad &
$\displaystyle 10. \quad \int x\sec^2(x^2)\,dx $
\vspace{0.15in}\\
$\displaystyle 11. \quad \int \dfrac{x}{x^2+1}\,dx $
& \qquad \qquad &
$\displaystyle 12. \quad \int \dfrac{1}{x^2+1}\,dx $
\vspace{0.15in}\\
$\displaystyle 13. \quad \int \frac{x+3}{x^2+6x-7}\,dx $
& \qquad \qquad &
$\displaystyle 14. \quad \int \frac{e^{1/x}}{x^2}\,dx $
\vspace{0.15in}\\
$\displaystyle 15. \quad \int \frac{\ln(x)}{x}\,dx $
& \qquad \qquad &
$\displaystyle 16. \quad \int e^{e^x}e^x\,dx $
\vspace{0.15in}\\
\end{tabular}
\end{center}

\noindent 
\force \qquad \qquad
17. \quad
Solve the initial value problem $y' = \cos(x)+e^x+1$ where $y(0)=3$.

\vspace{0.1in}

\noindent 
\force \qquad \qquad
18. \quad
Find the antiderivative $f(x)$ of 
$g(x) = \displaystyle{\frac{500}{\sqrt{2x+1}}+6x^2+3}$ such that
$f(4)=10$.


\vspace{0.25in}

{\bf\Large Answers:}

\begin{enumerate}
\item $\displaystyle{\int 3x^2-\frac{1}{x}+e^x-7\,dx = x^3-\ln|x|+e^x-7x+C}$

\vspace{0.05in}

\item $\displaystyle{\int \frac{\sqrt{x}}{x^3} + \frac{6+2x}{x^2}-50x^{99}\,dx 
       = \int x^{-3+1/2} + \frac{6}{x^2}+\frac{2x}{x^2}-50x^{99}\,dx }$ \\
      $\displaystyle{
       = \int x^{-5/2} + 6x^{-2}+\frac{2}{x}-50x^{99}\,dx
       = \frac{x^{-3/2}}{-3/2} + 6 \frac{x^{-1}}{-1} + 2\ln|x|-50\frac{x^{100}}{100}+C }$
      $\displaystyle{
       = -\frac{2}{3x^{3/2}} - \frac{6}{x} + 2\ln|x|-\frac{1}{2}x^{100}+C}$

\vspace{0.05in}

\item $\displaystyle \int 3\sin(x)+10\sec(x)\tan(x)\,dx  = -3\cos(x)+10\sec(x)+C$

\vspace{0.05in}

\item $\displaystyle \dfrac{5}{\sqrt{1-x^2}}\,dx = 5\,\mathrm{arcsin}(x)+C$

\vspace{0.05in}

\item $\displaystyle{\int \,dx = x+C}$

\vspace{0.05in}

\item $\displaystyle{\int \frac{e^{5x}}{e^{2x}} - 8x^{7/3}\,dx 
       = \int e^{5x-2x}-8x^{7/3}\,dx = \int e^{3x}-8x^{7/3}\,dx }$ 
      $\displaystyle{
       = \frac{1}{3}e^{3x} -8 \frac{x^{10/3}}{10/3}+C
       = \frac{e^{3x}}{3} - \frac{24}{10}x^{10/3}+C}$

\vspace{0.05in}

\item $\displaystyle{\int (x^2-5x+6)^{9}(2x-5)\,dx}$ \quad  Substitute $u=x^2-5x+6$ so that
      $du = (2x-5)\,dx$.\\ So our integral becomes 
      $\displaystyle{\int u^9\,du = \frac{u^{10}}{10}+C = \frac{(x^2-5x+6)^{10}}{10}}$

\vspace{0.05in}

\item $\displaystyle{\int e^{6x-2}\,dx}$ \quad  Substitute $u=6x-2$ so that
      $du = 6\,dx$ thus $(1/6)\,du = dx$.\\ So our integral becomes 
      $\displaystyle{\int e^u \frac{1}{6}\,du = \frac{1}{6}e^u+C = \frac{e^{6x-2}}{6}+C}$

\vspace{0.05in}

\item $\displaystyle \int \cos(5x+1)\,dx $ \quad  Substitute $u=5x+1$ so that $du=5\,dx$ thus $(1/5)\,du=dx$.\\
So our integral becomes 
$\displaystyle \int \cos(u)\frac{1}{5}\,du = \frac{1}{5}\sin(u)+C = \dfrac{1}{5}\sin(5x+1)+C$

\vspace{0.05in}

\item $\displaystyle \int x\sec^2(x^2)\,dx $ \quad Substitute $u=x^2$ so that $du=2x\,dx$ thus $(1/2)\,du=dx$.\\
So our integral becomes 
$\displaystyle \int \sec^2(u) \frac{1}{2}\,du = \frac{1}{2}\tan(u)+C = \frac{1}{2}\tan(x^2)+C$

\vspace{0.05in}

\item $\displaystyle \int \dfrac{x}{x^2+1}\,dx $ \quad Substitute $u=x^2+1$ so that $du=2x\,dx$ thus $(1/2)\,du=dx$\\
So our integral becomes 
$\displaystyle \int \dfrac{1}{u} \cdot \dfrac{1}{2}\,du = \dfrac{1}{2}\ln|u|+C = \dfrac{1}{2}\ln(x^2+1)+C$

\vspace{0.05in}

\item $\displaystyle \int \dfrac{1}{x^2+1}\,dx =\mathrm{arctan}(x)+C$ \quad [Just a basic formula here!]

\vspace{0.05in}

\item $\displaystyle{\int \frac{x+3}{x^2+6x-7}\,dx}$ \quad  Substitute $u=x^2+6x-7$ so that
      $du = (2x+6)\,dx = 2(x+3)\,dx$ thus $(1/2)\,du = (x+3)dx$.\\ So our integral becomes 
      $\displaystyle{\int \frac{(1/2)\,du}{u} = \frac{1}{2} \int \frac{1}{u}\,du
       = \frac{1}{2} \ln|u|+C}$  $\displaystyle{ = \frac{1}{2} \ln|x^2+6x-7|+C}$

\vspace{0.05in}

\item $\displaystyle{\int \frac{e^{1/x}}{x^2}\,dx = \int e^{x^{-1}}x^{-2}\,dx}$ \quad 
      Substitute $u=x^{-1}$ so that $du = -x^{-2}\,dx$ thus $-du = x^{-2}\,dx$.\\ 
      So our integral becomes $\displaystyle{\int e^u(-du) = -e^u+C = -e^{1/x}+C}$

\vspace{0.05in}

\item $\displaystyle{\int \frac{\ln(x)}{x}\,dx = \int \ln(x) \frac{1}{x}\,dx}$ \quad  
      Substitute $u=\ln|x|$ so that $du = (1/x)\,dx$.\\ So our integral becomes
      $\displaystyle{\int u\,du = \frac{1}{2}u^2+C = \frac{1}{2}(\ln|x|)^2+C}$

\vspace{0.05in}

\item $\displaystyle{\int e^{e^x}e^x\,dx}$ \quad Substitute $u=e^x$ so that
      $du = e^x\,dx$. So our integral becomes 
      $\displaystyle{\int e^u\,du = e^u+C = e^{e^x}+C}$

\vspace{0.05in}

\item Solve the initial value problem $y' = \cos(x)+e^x+1$ where $y(0)=3$.

$\displaystyle y = \int y'\,dx = \int \cos(x)+e^x+1\,dx = \sin(x)+e^x+x+C$.\\ We require $3=y(0)=\sin(0)+e^0+0+C = 1+C$ so $C=2$. Thus $y=\sin(x)+e^x+x+2$.

\vspace{0.05in}

\item Find the antiderivative $f(x)$ of 
$g(x) = \displaystyle{\frac{500}{\sqrt{2x+1}}+6x^2+3}$ such that
$f(4)=10$.

$\displaystyle{f(x) = \int g(x)\,dx = \int 500(2x+1)^{-1/2}+6x^2+3\,dx
 = 500(2x+1)^{1/2}+2x^3+3x+C}$ (For the term ``$500(2x+1)^{-1/2}$'' use the 
substitution $u=2x+1$). We also know that $f(4)=10$ so
$\displaystyle{10 = f(4) = 500(2(4)+1)^{1/2}+2(4^3)+3(4)+C}$
$10 = 500\sqrt{9}+128+12+C$ so $10 = 1500+140+C$ so $C = -1630$.

\vspace{0.05in}

Therefore, $f(x) = 500\sqrt{2x+1}+2x^3+3x-1630$
\end{enumerate}



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