$1+1+2$
$\frac{4}{7}x=-12$
$\:30\:\left(\frac{1}{2}\:x\:-\:2\right)\:+\:40\:\left(\frac{3}{4}\:y\:-\:4\right)$
$h\left(x\right)=\tan\left(\arccos\left(\frac{x-1}{x+2}\right)\right)$
$\int3x\cos\left(2x^2-3\right)dx$
$\int\left(\left(x^5+x^2\right)^7\left(5x^4+2x\right)\right)dx$
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