$36-9u=27$
$\left(\frac{-10}{2}\right)^2$
$\left(4x^{4}-9x^{2}+3x-2\right):\left(x^{2}+3x-5\right)$
$\int_0^{\frac{1}{3}}\arctan\left(3x\right)dx$
$-3n-8$
$1+\left(n+1\right)$
$\sec^4\left(u\right)-\sec^2\left(u\right)=\tan^2\left(u\right)+\tan^4\left(4\right)$
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