If $I\left( m \right) = \int\limits_0^\pi {\ln\left( {1 - 2m\cos x + {m^2}} \right)dx} $ then evaluate $\frac{I(9)}{I(3)}$

Question

If $I\left( m \right) = \int\limits_0^\pi {\ln\left( {1 - 2m\cos x + {m^2}} \right)dx} $ then evaluate $\frac{I(9)}{I(3)}$

My approach is as follow

$I\left( m \right) = \int\limits_0^\pi {\ln\left( {1 - 2m\cos x + {m^2}} \right)dx} $

$I\left( m \right) = \int\limits_0^\pi {\ln\left( {1 - 2m\cos \left( {\pi + 0 - x} \right) + {m^2}} \right)dx \Rightarrow } I\left( m \right) = \int\limits_0^\pi {\ln\left( {1 + 2m\cos x + {m^2}} \right)dx} $

$2I\left( m \right) = \int\limits_0^\pi {\left( {\ln\left( {1 - 2m\cos x + {m^2}} \right) + \ln\left( {1 + 2m\cos x + {m^2}} \right)} \right)dx} \Rightarrow 2I\left( m \right) = \int\limits_0^\pi {\left( {\ln\left( {{{\left( {1 + {m^2}} \right)}^2} - 4{m^2}{{\cos }^2}x} \right)} \right)dx} $

Not able to approach further