$\frac{8\tan^2\left(u\right)}{\sec^2\left(u\right)}$
$x+1<2$
$t^2-16t+8t$
$\:\frac{-56}{8}$
$\frac{y}{x}=u$
$\left(2+a\right)^3\left(2+a\right)^{\frac{1}{4}}\left(2+a\right)^{\frac{3}{2}}$
$a^2-25=0$
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