$\frac{1}{\left(1-7x\right)^3}$
$\int_0^{2\sqrt{3}}\left(\frac{z^3}{\sqrt{z^2+4}}\right)dz$
$\left(\sec^2\left(x\right)+1\right)\left(\sec^4\left(x\right)\right)+\cot^2\left(x\right)$
$e^y\left(1+y'\right)=0$
$y''-5y'-6y=0$
$x^2+2\cdot x\cdot5+5^2$
$x^4-4$
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