Answer
$$\int(2+\tan^2\theta)d\theta=\theta+\tan\theta+C$$
Work Step by Step
$$A=\int(2+\tan^2\theta)d\theta$$ $$A=\int[1+(1+\tan^2\theta)]d\theta$$ $$A=\int(1+\sec^2\theta)d\theta$$ From Table 1, $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ Therefore, $$A=\int(1)d\theta+\int(\sec^2\theta)d\theta$$
From Table 1, we also get the followings $$\int kdx=kx+C$$ $$\int(\sec^2 x)dx=\tan x+C$$
Therefore, $$A=\theta+\tan\theta+C$$
