If $\int\frac{x\,dx}{x^2-4x+8}=k\log(x^2-4x+8)+\tan^{-1}(\frac{x-2}{2})+C$, then the value of k is:

$\frac{1}{2}$ 1 2 None of these

$\frac{1}{2}$

Put x - 2 = u to get $I=\int\frac{x\,dx}{x^2-4x+8}=\int\frac{u+2}{u^2+4}du$ $=\frac{1}{2}\int\frac{2udu}{u^2+4}du+2\int\frac{du}{u^2+4}=\frac{1}{2}\log(u^2+4)+\tan^{-1}(\frac{u}{2})+C=\frac{1}{2}\log(x^2-4x+8)+\tan^{-1}(\frac{x-2}{2})+C$