$\int {\mathrm{Tan}}^{-1}\left(\frac{8x-4}{3-4x^2+4x}\right)dx=$

$$\begin{gathered} \begin{array}{l}\int {\tan }^{-1}\left(\frac{8x-4}{3-4x^2+4x}\right)dx=\int {\tan }^{-1}\left(\frac{4(2x-1)}{4-(2x-1{)}^2}\right)dx\\ =\int {\tan }^{-1}\left(\frac{(2x-1)}{1-{\left(\frac{2x-1}{2}\right)}^2}\right)dx\\ =\int 2{\tan }^{-1}\left(\frac{(2x-1)}{2}\right)dx \\ \left(\frac{2x-1}{2}\right)=t\Rightarrow dx =dt \\ =2\int {\tan }^{-1}tdt=2t {\tan }^{-1}t-\int \frac{2t}{1+t^2}dt\\ =2\left(\frac{2x-1}{2}\right) {\tan }^{-1}\left(\frac{2x-1}{2}\right)-\log \left(1+{\left(\frac{2x-1}{2}\right)}^2\right)+C \\ =(2x-1) {\tan }^{-1}\left(\frac{2x-1}{2}\right)-\log \left(4x^2-4x+5\right)+C \\ \end{array} \end{gathered}$$