Answer
$$\ln \left| {\frac{{\tan x}}{{\tan x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^2}x}}{{\tan x\left( {\tan x + 1} \right)}}} dx \cr
& {\text{Let }}t = \tan x,{\text{ }}dt = {\sec ^2}xdx \cr
& {\text{Substituting}} \cr
& \int {\frac{{{{\sec }^2}x}}{{\tan x\left( {\tan x + 1} \right)}}} dx = \int {\frac{1}{{t\left( {t + 1} \right)}}} dt \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{1}{{t\left( {t + 1} \right)}} = \frac{A}{t} + \frac{B}{{t + 1}} \cr
& 1 = A\left( {t + 1} \right) + Bt \cr
& t = 0 \to A = 1 \cr
& t = - 1 \to B = - 1 \cr
& \frac{1}{{t\left( {t + 1} \right)}} = \frac{1}{t} + \frac{{ - 1}}{{t + 1}} \cr
& \int {\frac{1}{{t\left( {t + 1} \right)}}} dt = \int {\left( {\frac{1}{t} + \frac{{ - 1}}{{t + 1}}} \right)} dt \cr
& {\text{Integrate}} \cr
& {\text{ = ln}}\left| t \right| - \ln \left| {t + 1} \right| + C \cr
& {\text{By using logarithmic properties}} \cr
& = \ln \left| {\frac{t}{{t + 1}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \ln \left| {\frac{{\tan x}}{{\tan x + 1}}} \right| + C \cr} $$
