Answer
$(\theta^2+1)^{3/2}+c$
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 $k=\theta^2+1 \implies 2\theta d\theta =dk$
Now, $\int 3 \theta \sqrt {\theta^2+1} d \theta=\dfrac{3}{2} \int (\theta^2+1)^2 (2\theta d\theta)=\dfrac{3}{2} \int k^{1/2} dk $
or, $=\dfrac{3}{2} [\dfrac{k^{1/2+1}}{1/2+1}]+c$
or, $=k^{3/2}+c$
plug back in: $k=\theta^2+1 $
Thus, $=(\theta^2+1)^{3/2}+c$
