$4\:x^3+12x^2+5x\:-6$
$x^2-kx+121=0$
$y'=y\left(4x-3\right)$
$y'=\frac{y}{t}$
$\int\frac{12y^2}{4y^3+1}dx$
$\int\left(y\csc\left(3y^2\right)\cot\left(3y^2\right)\right)dy$
$\sec\left(x\right)-\tan\left(x\right)\:\cdot\csc\left(x\right)+1$
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