$3\left(3x^2\right)^2\left(-1\right)$
$\int x\cdot\left(1+x\right)e^{1+x}dx$
$x^2\:+\:3x\:=\:24$
$\int\sec^{3}\thetad\theta$
$3x-2-1$
$_+5+-3$
$\frac{3x-5}{4}>3$
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