$\int_1^2\left(-\frac{4x}{y}\right)dx$
$16w^2-9z^2$
$\frac{dy}{dx}=sin\left(x\right)cos\left(x\right)$
$\frac{dy}{dx}=9xy^2$
$-\left(23-45\right)-\left(75-34\right)$
$\frac{dy}{dx}=ax^2+1$
$\int\frac{\left(x+3\right)}{\sqrt[3]{x^2+6x}}dx$
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