Evaluate $\int_0^1 \sqrt{x^2+4x+1}\,dx$ 
Evaluate $\displaystyle\int_0^1 \sqrt{x^2+4x+1}\,dx$

I tried simplify by doing this:
$$\int_0^1 \sqrt{x^2+4x+1}\:dx=\displaystyle\int_0^1 \sqrt{(x+2)^2-3}\:dx$$
Then, by letting $t=x+2$, $dt=dx$
$$\int_0^1 \sqrt{t^2-3}\:dt=\displaystyle\int_0^1 \sqrt{(t-\sqrt 3)(t+\sqrt 3)}\:dt$$
I tried solving $\displaystyle\int_0^1 \sqrt{t^2-3}\:dt$ by substituting: $$k=t^2-3, dt=\frac{dk}{2\sqrt{|k+3|}}$$
But that leads me nowhere:
$$\frac{1}{2}\displaystyle\int_0^1 \frac{\sqrt{k}}{\sqrt{|k+3|}}\:dk$$
Any hints?

HINT:
Set $x+2=\sqrt3\sec u\implies\sqrt3\tan u=\sqrt{x^2+4x+1}$
$\int\sqrt{(x+2)^2-3}dx=\int\sqrt3\tan u(\sqrt3\sec u\tan u)du=3\int(\sec^3u-\sec u)\ du$
See How to integrate $\sec^3 x \, dx$?
