$$ { =\frac{1}{2}\int{\sec xdx+\frac{1}{2}\int{\csc xdx-\frac{1}{2}\int{\frac{\frac{1}{\cos ^2x}}{\frac{\sin x\cos x}{\cos ^2x}}dx}}}} \\ { =\frac{1}{2}\ln \left| \sec x+\tan x \right|+\frac{1}{2}\ln \left| \csc x-\cot x \right|-\frac{1}{2}\int{\frac{\sec ^2x}{\tan x}dx}} \\ { =\frac{1}{2}\ln \left| \sec x+\tan x \right|+\frac{1}{2}\ln \left| \csc x-\cot x \right|-\frac{1}{2}\int{\frac{d\tan x}{\tan x}}} \\ { =\frac{1}{2}\ln \left| \sec x+\tan x \right|+\frac{1}{2}\ln \left| \csc x-\cot x \right|-\frac{1}{2}\ln \left| \tan x \right|+C} \\ { =\frac{1}{2}\left( \ln \left| \sec x+\tan x \right|+\ln \left| \csc x-\cot x \right|-\ln \left| \tan x \right| \right) +C.} \\ {\left( 1 \right) 解法2：\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}；1+\cos x=2\cos ^2\frac{x}{2}} \\ {原式=\int{\frac{2d\left( \frac{x}{2} \right)}{2\sin \frac{x}{2}\cos \frac{x}{2}+2\cos ^2\frac{x}{2}}=\int{\frac{d\left( \frac{x}{2} \right)}{\sin \frac{x}{2}\cos \frac{x}{2}+\cos ^2\frac{x}{2}}}}} \\ {=\int{\frac{\frac{1}{\cos ^2\frac{x}{2}}}{\frac{\sin \frac{x}{2}\cos \frac{x}{2}+\cos ^2\frac{x}{2}}{\cos ^2\frac{x}{2}}}d\left( \frac{x}{2} \right) =\int{\frac{\sec ^2\frac{x}{2}}{\tan \frac{x}{2}+1}d\left( \frac{x}{2} \right)}}=\int{\frac{d\left( \tan \frac{x}{2} \right)}{\tan \frac{x}{2}+1}}} \\ {=\int{\frac{d\left( \tan \frac{x}{2}+1 \right)}{\tan \frac{x}{2}+1}}=\ln \left| \tan \frac{x}{2}+1 \right|+C.} \\ { \left( 2 \right) 解：} \\ { 原式=\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{\left( \sqrt{1+x}+\sqrt{1-x}+\sqrt{2} \right) \left( \sqrt{1+x}+\sqrt{1-x}-\sqrt{2} \right)}}dx} \\ { =\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{\left( \sqrt{1+x}+\sqrt{1-x} \right) ^2-\left( \sqrt{2} \right) ^2}}dx=\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{\left( \sqrt{1+x}+\sqrt{1-x} \right) ^2-2}}dx} \\ { =\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{\left( \sqrt{1+x} \right) ^2+2\sqrt{1+x}\cdot \sqrt{1-x}+\left( \sqrt{1-x} \right) ^2-2}}dx} \\ { =\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{\left( 1+x \right) +2\sqrt{1+x}\cdot \sqrt{1-x}+\left( 1-x \right) -2}}dx} \\ { =\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{2\sqrt{1+x}\cdot \sqrt{1-x}+2-2}}dx} \\ { =\int{\frac{\sqrt{1+x}+\sqrt{1-x}-\sqrt{2}}{2\sqrt{1+x}\cdot \sqrt{1-x}}}dx} \\ { =\int{\frac{dx}{2\sqrt{1-x}}+\int{\frac{dx}{2\sqrt{1+x}}-\frac{\sqrt{2}}{2}\int{\frac{dx}{\sqrt{1+x}\cdot \sqrt{1-x}}}}}} \\ { =-\int{\frac{d\left( 1-x \right)}{2\sqrt{1-x}}}+\int{\frac{d\left( 1+x \right)}{2\sqrt{1+x}}-\frac{\sqrt{2}}{2}\int{\frac{dx}{\sqrt{1-x^2}}}}} \\ { =-\sqrt{1-x}+\sqrt{1+x}-\frac{\sqrt{2}}{2}\mathrm{arc}\sin x+C} \\ { =\sqrt{1+x}-\sqrt{1-x}-\frac{\sqrt{2}}{2}\mathrm{arc}\sin x+C.} $$