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. 

Using the substitution $u = 1 + t^2$, $du = 2t \,dt$, gives
$$\int 3t \sqrt{1 + t^2} dt = \frac{3}{2} \int \sqrt{u} \,du = u^{3/2} + C = (1 + t^2)^{\frac{3}{2}} + C,$$
and evaluating at the limits gives the identity.
Note that your substitution $t := \tan x$ works as well: We have
$$\int 3t \sqrt{1 + t^2} dt = 3\int \tan t \sec t \cdot \sec^2 t \,dt,$$ and we can substitute $v = \sec t$, $dv = \sec t \tan t$, but note that up to an overall power this achieves the same effect that the single substitution $u$ does.
