$x+6=-\frac{1}{2}x$

$\int\frac{u^2+2u-5}{u^3-u^2-4u+4}du$

$\frac{\left(\tan\left(x\right)+1+\sec\left(x\right)\right)\left(\tan\left(x\right)+1-\sec\left(x\right)\right)}{\tan\left(x\right)}$

$\left(x+3\right)\cdot\left(x-5\right)=0$

$-\int\frac{2}{v\left(v^2+1\right)}dv$

$5x^2+x-1.3x+2$

$e^{x^{2+6}}$