Answer
$$\frac{1}{2}\theta^2+\tan\theta +C$$
Work Step by Step
$\int(\theta+\sec^2\theta)d\theta$
Given $$\int{x}dx=\frac{1}{2}x^2+C$$ and$$\int\sec^2{x}dx=\tan{x}+C$$
Thus,
$\int(\theta+\sec^2\theta)d\theta$
$=\frac{1}{2}\theta^2+\tan\theta +C$
