Math Through Answer

$\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}dx=\frac{22}{7}-\pi$
${ x + a = a + x = x}$

$$
\frac{x^n-1}{x-1} = \sum_{k=0}^{n-1}x^k
$$

$$
\frac{a/b-c/d}{e/f-g/h}
$$

$\frac{1}{2}$
$\frac{2}{x+2}$

\documentclass{article}
\begin{document}
%You won't see this in the final document.
You do see this.
\end{document}

Evaluate the sum $\displaystyle\sum\limits_{i=0}^n i^3$.

$
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\
&= 2x^2 + 3x^2 - 9x + 6\\
&= 5x^2 - 9x + 6
$


$\frac {x^2+5}{3x +1}$

$$
\sum_{k=0}^\infty\frac{(-1)^k}{k+1} = \int_0^1\frac{dx}{1+x}
$$

$$
50 apples \times 100 apples = lots of apples
$$

\begin{eqnarray*}
1+2+\ldots+n&=&\frac{1}{2}((1+2+\ldots+n)+(n+\ldots+2+1))\\
&=&\frac{1}{2}\underbrace{(n+1)+(n+1)+\ldots+(n+1)}_{\mbox{$n$copies}}\\
&=&\frac{n(n+1)}{2}\\
\end{eqnarray*}

\begin{document}
\begin{align*}
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\
&= 2x^2 + 3x^2 - 9x + 6\\
&= 5x^2 - 9x + 6
\end{align*}
\end{document}

$
\begin{align*}
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\
&= 2x^2 + 3x^2 - 9x + 6\\
&= 5x^2 - 9x + 6
\end{align*}
$

----------------

$\begin{align*}
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\
&= 2x^2 + 3x^2 - 9x + 6\\
&= 5x^2 - 9x + 6
\end{align*}$