Proving function of Brownian motion is a martingale

Question

In a question we're given, are given $X_t= \int_0^t sB_s dB_s$ for $B_s$ being Brownian motion. We are first told to find $\text{Var}(X_t)$, which I compute using Ito's Lemma to be $\frac{t^4}{4}$. Then, we seek to show in the second part that $$\sqrt{B_t^2 + 10} - 5\int_0^t \frac{ds}{(B_s^2 + 10)^{3/2}}$$ is a martingale. I mentioned the earlier part just in case it's relevant to proving this martingale fact, which is what we ultimately seek to prove. I think proving finite absolute means is doable, but I'm not sure how to proceed in showing the martingale property. I tried using linearity to split into two conditional expectations, and using Ito's Lemma on the second, but not sure where to go from there or how to make that work. In particular, I don't know how to rewrite the above functions of $B_t$ as a function of $B_s$ for some $s < t$.

Answer 1

StratosFair's suggestion is a good one.

Let $Y_t = \sqrt{B_t^2 + 10}$ so that, by Ito's formula, after a bit of rearranging, $dY_t = B_tY_t^{-1}dB_t + 5Y_t^{-3}dt$.

Then $dZ_t = B_tY_t^{-1}dB_t$. We know that an Ito integral against a Brownian is a (local) martingale, so we are done.