$r^4\:\left(-5\right)r^3$
$dy=4dx$
$-3+-4-2+-7-4-7+3$
$\sin^3x+sinxcos^2x=sinx$
$\sqrt{18a^4}$
$\int_{-\infty\:}^0\left(\frac{1}{4+x^2}\right)dx$
$1-7\left(4x+1\right)=6\left(-2x-3\right)-4$
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