$\int\frac{8+3x}{1+x^2}dx$
$x+68\ge75$
$y\:=\:\frac{sin^4x\cdot tan^6x}{\left(x^2+3\right)^2}$
$\left(-10\right)\cdot\left(-10\right)^3$
$\int\frac{5}{v^2+3v-4}dv$
$3\cdot x^6\cdot2y$
$\frac{dy}{dx}=\frac{4+x+y}{8+x+y}$
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