How to $\int 3t \sqrt {t^2+1} dt$ $$\int 3t \sqrt {t^2+1} dt$$
My book has done
$$\int_0^2 3t (1 + t^2)^{\frac{1}{2}}dt = \left[(1 + t^2)^{\frac{3}{2}}\right]_0^2.$$
They have done so simple way, but how is this correct? I have been taught that when $t$ (inside bracket) has a power greater than $1$, this isn't possible. I used the substitution $t := \tan x$, but I can't complete it. 

Substitute $u = t^2 + 1$. So $\frac{du}{dt} = 2t$.
$$
\int 3t \sqrt {t^2+1} dt = \frac{3}{2}\int \sqrt{u} \,du = \frac{3}{2} \times \frac{u\sqrt{u}}{\frac{3}{2}}+C=(1+t^2)^{\frac{3}{2}}+C.
$$
