I am interested in solving the following SDE: $$dX_t = \left( \sqrt{1+X_t^2} + \frac{X_t}{2}\right)dt + \sqrt{1+X_t^2} dW_t$$ but I am having quite a bit of trouble. This is Exercise 5.2.27 in Karatzas and Shreve, not a HW question, and is just for self study. I imagine I should be looking for a nice transformation such us the one used to solve the Ornstein Uhlenbeck SDE. Trying $Y_t = \log\left(\sqrt{1+X_t^2}\right)$ gives $$dY_t = \frac{X_tdX_t}{1+X_t^2} + \frac{(1-X_t^2)/2}{1+X_t^2}dt = \frac{X_t\sqrt{1+X_t^2} + X_t^2/2 + 1/2 - X_t^2/2}{1+X_t^2} dt + \frac{X_t}{\sqrt{1+X_t^2}}dW_t \\ = X_t e^{-Y_t} \left(dt + dW_t \right) + (1/2)e^{-2Y_t}dt$$
but for obvious reasons I feel like this is completely on the wrong track. Any help would be massively appreciated!
As hinted in the comments, with a term $\sqrt{1+x^2}$ that has no immediate resolution, one next step to try is to set $x=\sinh(y)$, so that $\sqrt{1+x^2}=\cosh(y)$. Then the SDE reads as $$ dX_t=(\cosh(Y_t)+\tfrac12\sinh(Y_t))dt+\cosh(Y_t)dW_t $$ On the other hand, the Ito formula for that substitution is $$ dX_t=\tfrac12\sinh(Y_t)\,d\langle Y_t\rangle+\cosh(Y_t)dY_t $$ Comparing both it is possible to match all terms, which results in $$ dY_t=dt+dW_t\implies Y_t=Y_0+t+W_t $$