Answer
$$\ln \left| {\frac{{\tan x + 2}}{{\tan x + 3}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^2}x}}{{{{\tan }^2}x + 5\tan x + 6}}} dx \cr
& {\text{Let }}t = \tan x,{\text{ }}dt = {\sec ^2}xdx \cr
& {\text{Substituting}} \cr
& \int {\frac{{{{\sec }^2}x}}{{{{\tan }^2}x + 5\tan x + 6}}} dx = \int {\frac{1}{{{t^2} + 5t + 6}}} dt \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{1}{{{t^2} + 5t + 6}} = \frac{1}{{\left( {t + 3} \right)\left( {t + 2} \right)}} = \frac{A}{{t + 3}} + \frac{B}{{t + 2}} \cr
& 1 = A\left( {t + 2} \right) + B\left( {t + 3} \right) \cr
& t = - 3 \to A = - 1 \cr
& t = - 2 \to B = 1 \cr
& \frac{1}{{{t^2} + 5t + 6}} = \frac{{ - 1}}{{t + 3}} + \frac{1}{{t + 2}} \cr
& \int {\frac{1}{{{t^2} + 5t + 6}}} dt = \int {\left( {\frac{{ - 1}}{{t + 3}} + \frac{1}{{t + 2}}} \right)} dt \cr
& {\text{Integrate}} \cr
& {\text{ = }} - {\text{ln}}\left| {t + 3} \right| + \ln \left| {t + 2} \right| + C \cr
& {\text{By using logarithmic properties}} \cr
& = \ln \left| {\frac{{t + 2}}{{t + 3}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \ln \left| {\frac{{\tan x + 2}}{{\tan x + 3}}} \right| + C \cr} $$
