Answer
\[\ln |\sin x\cos x+1|+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{\sec x\,\cos 2x}{\sin x+\sec x}\:dx\]
\[I=\int\frac{\sec x\,\cos 2x}{\sin x+\frac{1}{\cos x}}\:dx\]
\[I=\int\frac{\cos 2x}{\sin x\:\cos x+1}\:dx\]
Put $t=\sin x\:\cos x+1\;\;\;...(1)$
\[\Rightarrow dt=(\cos^2 x-\sin^2 x)dx=\cos 2x \:dx\]
\[I=\int\frac{dt}{t}=\ln|t|+C\]
Where $C$ is constant of integration
Using (1)
\[I=\ln |\sin x\cos x+1|+C\]
Hence, \[\int\frac{\sec x\,\cos 2x}{\sin x+\sec x}=\ln |\sin x\cos x+1|+C\]