$tx-4=-8$

$\int\frac{x^2-3}{x\left(x+1\right)\left(x-3\right)}dx$

$8x\tan\left(2x\right)+8x^2\sec\left(2x\right)^2-6\sec\left(2x\right)^2$

$\int\left(e^{3x}\left(6+e^{3x}\right)^6\right)dx$

$\lim_{x\to0}\left(5x^2+3x-2\right)$

$b^2+b-1-8+2+7b+9u+6u+1-7b^2-3b+8b$

$5.8^2+5^{-2}$