$\int_1^{\infty}\left(\frac{x}{\left(bx^2+1\right)^3}\right)dx$
$\sin^210x$
$8x^2y^3\cdot x^4$
$3a-2b=8$
$\sqrt{169m^6n^8}$
$= \frac { a ^ { m + 8 } } { a ^ { m + 3 } }$
$\frac{d}{dx}\left(\sec\left(xy\right)\ln\left(x^y\right)=5\right)$
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