Official solution
Let, \[{{{x - 4} \over {x + 2}}}\] = t
\[ \Rightarrow \] x - 4 = t(x+2)
\[ \Rightarrow \] x (1 -t) = 2(t+2)
\[ \Rightarrow \] x = \[{{2(t + 2)} \over {1 - t}}\]
\[ \therefore \] f(t) = 2(\[{{2(t + 2)} \over {1 - t}}\]) + 1
= \[{{4t + 8} \over {1 - t}} + 1\]
= \[{{3t + 9} \over {1 - t}}\]
= \[{{3(t + 3)} \over {1 - t}}\]
= \[{{3(t - 1 + 4)} \over {1 - t}}\]
= - 3 + \[12 \over 1 - t\]
\[ \therefore \] f(x) = - 3 + \[12 \over 1 - x\]
\[ \therefore \] \[\int {f(x)dx} \]
= \[\int {\left( { - 3 + {{12} \over {1 - x}}} \right)dx} \]
= \[-12{\log _e}|1 - x|\] - 3x + C