$5\left(x+1\right)-x\left(7-x\right)>x^2$

$\left(x+3\right)^2=x$

$\int_0^{\infty}\left(\frac{\left(\frac{2}{3}\right)x}{x^2+1}\right)dx$

$\lim_{x\to\infty}\left(\frac{2e^x}{1+e^x}\right)$

$8c-4-2c+5$

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

$\int\left(x^2+6\right)e^xdx$