Finding the expected value of of $\int_0^s \sqrt{t+B_t^2}dB_t$? How can I find the expected value of $\int_0^s \sqrt{t+B_t^2}dB_t$? I know one condition is to show that if:
$f: (0,\infty) \times \Omega \to \mathbb{R}$ is progressively measurable and $$\mathbb{E} \left( \int_0^s |f(t)|^2 \, dt \right)<\infty \quad \text{for all $s \geq 0$}$$ then
$$M_s := \int_0^s f(t) \, dB_t, \qquad s \geq 0,$$
is a martingale. 
However, I don't know how to prove the above nor know of any good names for it. Is there a way to do it by Ito's formula for space and time or any other direct methods? Thanks.

I think you already wrote down everything you need to solve your problem. 
About measurability. I assume that we are speaking about the filtration $\mathcal F_t$ that is generated by the given Brownian motion $B_t$. Then $f(t)=\sqrt{t+B_t^2}$ is progressively measurable with respect to $\mathcal F_t$. Indeed, $B_t$ itself is progressively measurable w.r.t. $\mathcal F_t$ (it is adapted to $\mathcal F_t$ by the definition of this filtration and has a.s. continuous paths). The function $g(t)=t$ is also progressively measurable w.r.t. $\mathcal F_t$ (it is non-random). Since $f(t)$ is a measurable function of $B_t$ and $g(t)$, it is progressively measurable as well.   
About expectation. We have $$\mathbb E \left(\int_0^s|f(t)|^2 dt\right)=\mathbb E \left(\int_0^st+B_t^2 dt\right)=\mathbb E \left(\frac{s^2}{2}+\int_0^sB_t^2 dt\right)=\frac{s^2}{2}+\mathbb E \left(\int_0^sB_t^2 dt\right).$$
Now by Tonelli's theorem, we may interchange expectation and integration in the last term:
$$\mathbb E \left(\int_0^sB_t^2 \ dt\right)=\int_0^s\mathbb E(B_t^2)\ dt=\int_0^s t \ dt=\frac{s^2}{2}.$$ 
We obtain that
$$\mathbb E \left(\int_0^s|f(t)|^2 dt\right)=s^2<\infty.$$
Now we conclude that your integral is a martingale, and its expected value is therefore zero. 
