Answer
$\sqrt {7+\theta^2}+c$
Work Step by Step
Since, we know $\int x^{a} dx=\dfrac{x^{a+1}}{n+1}+c$
where $c$ is a constant of proportionality.
Plug $7+\theta^2=a $ and $(2\theta) d\theta =da$
Now, we have $\dfrac{1}{2} \int \dfrac{1}{\sqrt a} da=(\dfrac{1}{2}) \int a^{-1/2} da$
and $(\dfrac{1}{2}) \int a^{-1/2} da=(\dfrac{1}{2}) [\dfrac{a^{-1/2+1}}{-\frac{1}{2}+1}]+c=a^{(1/2)}+c$
Hence, $a^{(1/2)}+c=\sqrt {7+\theta^2}+c$
