$8\left(\frac{1}{4}\right)+3\left(1\right)\frac{1}{3}$
$\int x ^ { - 6 } d x$
$\frac{x+7}{x^2-x-6}$
$-4y+8-3x+4y+9$
$\int\left(\frac{9\left(5+7tan^{-1}x\right)^3}{1+x^2}\right)dx$
$\frac{d}{dx}y=\sqrt[3]{x^3+3x+4}.x^5$
$9x^2y+6xz-4xy-2xz+3x^2y$
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