$\left(a^3-3\right)\left(a^3+9\right)$
$\frac{3}{sec\:x-tan\:x}$
$\int\left(18x^2\left(3x^3+3\right)^2\right)dx$
$dw=\ln\left(\frac{uv}{st}\right)$
$\lim_{x\to\infty}\left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}}$
$x^2dx=-y\left(x-1\right)dy$
$3x+2\left(5x-7\right)$
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