$\frac{sec\:t-1}{tan\:t}=\frac{\left(tan\:t\right)}{sec\:t+1}$
$\int4x\left(11x^2-10x\right)dx$
$\int_{0}^{\ln3}\frac{e^{t}}{49+e^{2t}}dt$
$3x^2-x-9$
$b^4y^8-441c^{10}$
$\frac{5}{x-1}-\frac{1}{x+1}=1$
$\left(u^3\right)^3$
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