You submitted the following TeX

\documentstyle{article}
\pagestyle{empty}
\topmargin=-0.5in
\textheight=9in
\begin{document}

\begin{center}
{\large\bf Sample Equations} 
\end{center}

\vspace{.2in}


\begin{equation}
x' + y^{2} = z_{i}^{2}
\end{equation}

\vspace{.1in}

\begin{equation}
-2 \, \ln |x-2| + 3 \, \ln |x-3| + C
\end{equation}

\vspace{.1in}

\begin{equation}
y = k \, e^{x^2/2} - 1\;\; 
\end{equation}

\vspace{.1in}

\begin{equation}
 \frac{dr}{dt} = \frac{1}{\pi}
\end{equation}
\begin{equation}
[-1/\sqrt{3},0] \cup [1/\sqrt{3},+\infty[ 
\end{equation}

\vspace{.1in}

\begin{equation}
 \frac{\cos x}{2 \sqrt{\sin x}}
\end{equation}

\vspace{.1in}

\begin{equation}
 A = \int_0^1 \frac{\ln(x+1) \,
\sqrt{x^2 + 2x + 2}}{x + 1} \, dx
\end{equation}

\vspace{.1in}

\begin{equation}
 \frac{1}{2} \int_0^{\pi/3}
\sin^2 3\theta \; d\theta = \frac{\pi}{12}
\end{equation}

\end{document}


\vspace{.1in}


TtHMML translated this into:

Sample Equations


x'+ y 2 = zi 2 (1)


-2lnx-2+3lnx-3+C (2)


y=ke x 2 /2 -1     (3)


dr dt = 1 π (4)

[-1/3,0][1/3,+[ (5)


cosx 2sinx (6)


A=0 1 ln(x+1) x 2 +2x+2 x+1 dx (7)


1 2 0 π/3 sin2 3θ  dθ= π 12 (8)

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