$\frac{6\cdot2+12}{-3+2}$
$\int\:\left(6x^3+2\sqrt[3]{x^4}+1\right)dx$
$\sec^2\left(a\right)+tan^2\left(a\right)=2tan^2\left(a\right)+1$
$2x^2-6x+1=x^2-5x+3$
$y'=x^2y^{-5}$
$-t^2+8t+12$
$i x ^ { 2 } - x y + x ^ { 3 } - 8 = 0$
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