Denote by $\mu_x^c$ and $\mu_x^{\lambda}$ the measures induced by the process $x^c$ and $x^{\lambda}$ generated by the following SDE: $$dx^c(r) = cx^c(r) dr + dW(r)$$ $$dx^{\lambda}(r) = \lambda x^{\lambda}(r) dr + dW(r)$$ where $r \in (0,1)$. It can be seen that $\mu_x^c$ and $\mu_x^{\lambda}$ are equivalent, how to derive the following Radon-Nikodym derivative evaluated with $x^{\lambda}$?
$$\dfrac{d\mu_x^{c}}{d\mu_x^\lambda}(x^{\lambda}) = \exp\left\lbrace (c - \lambda)\int_{0}^{1} x^{\lambda}(r) dx^{\lambda}(r) - \dfrac{1}{2}(c^2 - \lambda^2)\int_{0}^{1} \left( x^{\lambda}(r)\right)^2 dr \right\rbrace$$
What if SDE has been changed to be $$dx^c(r) = c_1x^c(r) dr + c_2x^c(r) dB(r) + dW(r)$$ $$dx^{\lambda}(r) = \lambda_1 x^{\lambda}(r) dr + \lambda_2 x^{\lambda}(r) dB(r) + dW(r)$$ where $B(r)$ and $W(r)$ are independent Brownian motions?