$-\frac{31}{\left(-31\right)}$

$\int\left(\frac{\left(1-4x\right)}{\left(4x^2-2x+6\right)}\right)dx$

$2x^2+10x+7=0$

$x+20=4x-4$

$\frac{25}{4}+\frac{16x^2}{36}-\frac{10}{3}x$

$9\:2\sqrt{3\:}\:2\sqrt{3}$

$\csc\left(x\right)^2\tan\left(2x\right)=\frac{2\cos\left(x\right)}{\cos\left(2x\right)\sin\left(x\right)}$