$h=\cos\left(x\right)\cdot\sin\left(x\right)$
$\int\frac{\left(2x+4\right)}{x^3-2x^2}dx$
$\left(6a+10b\right)\left(6a-10b\right)$
$\int\left(x^2\cos\left(8-x^3\right)\right)dx$
$\frac{y'}{y\left(x+1\right)}=1$
$v^2-6v$
$8x-8=-4x+16$
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