$\left(xy-4x\right)dx=\left(x^2y+y\right)dy$
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}16\sec^4xdx$
$\frac{26}{0}$
$\left(8-y^3\right)\left(8+y^3\right)$
$x^2-4y^2-8=0$
$[\left(x+y\right)^{2}+\left(x-a\right)]^{2}$
$-4\left(9\right)^3+6\left(9\right)^2+\left(9\right)-2$
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