## Calculus (3rd Edition)

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