Answer
$\sqrt {7+\theta^2}+c$
Work Step by Step
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$
where $c$ is a constant of proportionality.
Substitute $k=7+\theta^2 \implies 2\theta d\theta =dk$
Thus,
$\dfrac{1}{2} \int \dfrac{1}{\sqrt k} dk=\dfrac{1}{2} \int k^{-1/2} dk$
or, $=\dfrac{1}{2} [\dfrac{k^{-1/2+1}}{-1/2+1}]+c$
or, $=k^{1/2}+c$
plug back in: $k=7+\theta^2 $, we have
$=\sqrt {7+\theta^2}+c$