$$\int\left(1+tanx\right)^3sec^2xdx$$

$$\:\int\:\frac{1}{\left(1+121x^2\right)^{\frac{3}{2}}}dx$$

$$\frac{\tan\left(1\right)}{\sec\left(1\right)}=\sin\left(1\right)$$

$$\left(x+03\right)^2$$

$$\sec^2x=3\tan x-1$$

$$-7x-2-9y+10x$$

$$u+v=-13$$