$\left(-\frac{4}{3}x^2+\frac{12}{5}x-6\right)\cdot\left(\frac{3}{4}x^4\right)$
$-4\left(-6\right)-11$
$\frac{dy}{dx}\left(2x+3y\right)$
$x^2y'=\left(x+1\right)y$
$\int\left(\frac{1}{x^5\sqrt{4x^2-1}}\right)dx$
$3x^5-2x^4+2x^2+4-1$
$6x+8=10x-4$
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