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$\eqalign{ & \int {\frac{1}{x}{\kern 1pt} {\text{d}}x} \cr & x > 0, \cr & = \ln x + C \cr & \# \left( {\ln x + C} \right)' = \frac{1}{x}\# \cr & x < 0, \cr & = \ln ( - x) + C \cr & \# \left[ {\ln ( - x) + C} \right]' = \frac{1}{{ - x}} \times ( - 1) = \frac{1}{x}\# \cr & \therefore \cr & = \ln |x| + C \cr & \# \# \cr & \int {\frac{{\cos x}}{{\sin x + \cos x}}{\kern 1pt} {\text{d}}x} \cr & = \int {\left[ {\frac{{\cos x - \sin x}}{{2\left( {\sin x + \cos x} \right)}} + \frac{1}{2}} \right]{\text{d}}x} \cr & = \frac{1}{2}x - \frac{1}{2}\int {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}{\kern 1pt} {\text{d}}x} \cr & u = \sin x + \cos x,{\text{d}}x = \frac{1}{{\cos x - \sin x}}{\kern 1pt} {\text{d}}u, \cr & = \frac{1}{2}x + \frac{1}{2}\int {\frac{1}{u}{\kern 1pt} {\text{d}}u} \cr & = \frac{1}{2}x + \frac{1}{2}\ln u \cr & = \frac{x}{2} + \frac{{\ln \left( {\sin x + \cos x} \right)}}{2} \cr & = \frac{{\ln \left| {\sin x + \cos x} \right| + x}}{2} + C \cr}$

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