To finish the job initiated by Sivaram Ambikasaran, first you can say that $\mu(x)=\alpha x+ \beta 1/x$
The stochastic integral term with $\beta=0$ case is answered by Sivaram Ambikasaran.
For the $\alpha=0$ case you have using Itô's lemma :
$dLnB_t= \frac{1}{B_t}dB_t-\frac{1}{2.B_t^2}dt$
So integrating this gives you :
$\beta.\int_0^T \frac{dB_s}{B_s}=\beta.(Ln(\frac{B_T}{B_0})+1/2\int_0^T \frac{1}{B_s^2}ds)$
If you start your Brownian Motion at 0, the problem is not well defined. If $B_0\not=0$ then returning to the intial problem (still with $\alpha=0$) you have :
$ \int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2} =
Ln(\frac{B_T}{B_0})+\frac{\beta.(1-\beta)}{2}.\int_0^T \frac{1}{B_s^2}ds$ if I am not mistaken.
For the general case ($\alpha\not= 0$ and $\beta\not= 0$) just adds the solutions ( i.e. respectively for $\alpha=0$ and $\beta=0$) and also add the cross product term from $\int_0^T (\mu(B_s))^2 ds$ (which should be something like $-T.\alpha.\beta$).
Regards