$\int\frac{x^2+4x-1}{\left(x-1\right)\left(x-2\right)\left(x+3\right)}dx$
$\int\left(tan^34xsec^5\left(4x\right)\right)dx$
$\frac{5x^3-3}{x^2+2}$
$\frac{dy}{dt}=ln\left(3t\right)y$
$x^2\left(3y^3\right)x\:2\:$
$18x+5-10$
$\frac{d}{dx}y^2=9x$
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