Simplifying
$\left(t-\frac{1}{t}\right)^2$
$5x^3+x^2-5x-1$
$\lim_{x\to\infty}\left(\frac{\sin\left(e^x\right)}{x^2}\right)$
$tan^2x-sec^2x=-1$
$169+b^2-26b$
$\int_0^{\infty}\left(\frac{1}{\left(x^2+1\right)^3}\right)dx$
$\frac{dz}{dx}=\frac{-19z^2}{3}$
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