How find stochastic logarithm of $B^2(t)+1$. Find the stochastic logarithm of $B^2(t)+1$.
I know that for find stochastic logarithm According to Theorem we must use the The following formula $$X(t)=\mathcal L(U)(t)= \ln(\frac{U(t)}{U(0)})+\int_{0}^{t} \frac{d[U,U](t)}{2U^2(t)} $$  
I start and if $U(t)=B^2(t)+1$ Then $dU(t)=dt+2B(t)dB(t)$ and $d[U,U](t)=4 B^2(t)dt$ 
and my problem is I cannot get a close form or I cannot Calculate the integral. 
thanks for help
 A: You might be expected to note that
$$
X_t=\int_0^t\frac{\mathrm dU_s}{U_s}=2\int_0^t\frac{B_s}{B_s^2+1}\mathrm dB_s+\int_0^t\frac{\mathrm ds}{B_s^2+1}.
$$
A: I happen to meet this problem as Exercise 5.8 on Fima C. Klebaner's Introduction to Stochastic Calculus with Applications (3ed).
Just want to make an observation: the two results -- computed directly by definition and by the formula mentioned in the question -- are indeed the same.
On one hand, we may directly compute by definition. That is Did's answer:
$$X(t)=\mathcal L(U)(t)=\int_0^t\frac{\mathrm dU(s)}{U(s)}=\int_0^t\frac{2B(s)}{B(s)^2+1}\mathrm dB(s)+\int_0^t\frac{1}{B(s)^2+1}\mathrm ds$$
On the other hand, we may use the formula mentioned in the question, which in this case yields
$$
X(t)=\mathcal L(U)(t)= \ln(B^2(t)+1) \ +\int_{0}^{t} \frac{2B^2(s)}{[B^2(s)+1]^2}\mathrm ds
$$
The two results are indeed the same, which can be seen if we take the differential of $\ln(B^2(t)+1)$ by Ito's formula, which is
$$
\mathrm d\ln(B^2(t)+1) = \frac{2B(s)}{B^2(s)+1}\mathrm dB(s) + \frac{1-B^2(s)}{[B^2(s)+1]^2}\mathrm ds
$$
Note that after combining the $\mathrm ds$ terms,
$$
\frac{1-B^2(s)}{[B^2(s)+1]^2} + \frac{2B^2(s)}{[B^2(s)+1]^2} = \frac{B^2(s)+1}{[B^2(s)+1]^2} =  \frac{1}{B^2(s)+1} 
$$
So both the terms on $\mathrm dB(s)$ and $\mathrm ds$ are the same for the two formulae.
