Answer
$(\theta^2+1)^{(3/2)}+c$
Work Step by Step
Use formula $\int x^{a} dx=\dfrac{x^{(a+1)}}{(a+1)}+c$
where $c$ refers to a constant of proportionality.
Plug $\theta^2+1=a $ and $2\theta d\theta =da$
Now, $\int 3 \theta \sqrt {\theta^2+1} d \theta=\dfrac{3}{2} \int (\theta^2+1)^2 (2\theta d\theta) $
and $\dfrac{3}{2} \int a^{1/2} dk=\dfrac{3}{2} [\dfrac{a^{\frac{1}{2}+1}}{\frac{1}{2}+1}]+c=a^{(3/2)}+c$
Hence, $a^{(3/2)}+c=(\theta^2+1)^{(3/2)}+c$
