example 0
0.First
\[\begin{aligned} \int\frac{1}{\sin x+\cos x}dx=? \\ \\ 设:u=\tan\frac{x}{2}, \enspace x=2\arctan(u) \\ \\ \sin x=\frac{2u}{1+u^{2}}, \enspace \cos x=\frac{1-u^{2}}{1+u^{2}} \\ \\ \int\frac{1}{\frac{2u}{1+u^{2}} +\frac{1-u^{2}}{1+u^{2}}}d\left(2\arctan u\right) \\ \\ d(2\arctan u)=du\cdot(2\arctan u)' \\ \\ [2\arctan(u)]^{\prime}=(2)^{\prime}\arctan(u)+2[\arctan(u)]'= \frac{2}{1+u^{2}} \\ \\ \int\frac{1+u^{2}}{2u+1-u^{2}}\cdot\frac{2}{1+u^{2}}du=2\int\frac{1}{1-u^{2}+2u}du \end{aligned} \]0.Second
\[\begin{aligned} 1-u^{2}+2u\Rightarrow1+(u^{2}+2u)-1+1 \\ \\ 1+(-u^{2}+2u-1)+1 \\ \\ \boxed {2+(u^{2}-2u+1)=2-(u-1)^{2} } \\ \\ \boxed {[\sqrt{2}-(u-1)][\sqrt{2}+(u-1)]} \\ \\ \Rightarrow 2\int\frac{\left[\sqrt{2}+\left(u-1\right)\right]}{\left[\sqrt{2}-\left(u-1\right)\right]\left[\sqrt{2}+\left(u-1\right)\right]}du \end{aligned} \]0.Thirdly
\[\begin{align} 设: u-1=j \\ \\ \frac{A(\sqrt{2}+j)}{(\sqrt{2}-j)(\sqrt{2}+j)}+ \frac{B(\sqrt{2}-j)}{(\sqrt{2}-j)(\sqrt{2}+j)} \\ \\ =\frac{A}{\sqrt{2}-j}+\frac{B}{\sqrt{2}+j} \\ \\ =A(\sqrt{2}+j)+B(\sqrt{2}-j)=1 \\ \\ A\sqrt{2}+Aj+B\sqrt{2}-Bj=1 \\ \\ \sqrt{2}(A+B)+j(A-B)=1 \\ \\ \because j 为未知变量 \\ \\ \begin{cases}A-B=0\Rightarrow A=B \\ \\ \sqrt{2}(A+B)\end{cases} \\ \\ 2\sqrt{2}A=1\Rightarrow A=B=\frac{1}{2\sqrt{2}} \\ \\ 2\int\left(\frac{\frac{1}{2\sqrt{2}}}{\sqrt{2}-(u-1)}+\frac{\frac{1}{2\sqrt{2}}}{\sqrt{2}+(u-1)}\right)du \\ \\ 2\int \frac{1}{2\sqrt{2}} [\frac{1}{\sqrt{2}-(u-1)}+\frac{1}{\sqrt{2}+(u-1)}] du \\ \\ \frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2}-(u-1)}du+\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2}+(u-1)}du \end{align} \]0.Fourth
\[\begin{align} 设:\sqrt{2}-(u-1)=h \\ \\ \int\frac{1}{\sqrt{2}-(u+1)}du=\int\frac{1}{h}d\left[\sqrt{2}-(u-1)\right] \\ \\ =\frac{1}{\sqrt{2}}\int\frac{1}{h}\cdot -1 dh=-\frac{1}{\sqrt{2}}\int\frac{1}{h}dh \\ \\ =-\frac{1}{\sqrt{2}}\ln\left|h\right|+C \\ \\ \boxed{ \frac{1}{\sqrt{2}}\ln\left|\sqrt{2}-\left(u-1\right)\right|+C } \\ \\ \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{2}+(u-1)} du \\ \\ \boxed{ =\frac{1}{\sqrt{2}} \ln |\sqrt{2}+(u-1)|+C } \end{align} \]0.Fifth
\[\begin{align} -\frac{1}{\sqrt{2}} \ln |\sqrt{2}-(u-1)|+\frac{1}{\sqrt{2}} \ln |\sqrt{2}+(u-1)| \\ \\ \frac{1}{\sqrt{2}} \ln |\sqrt{2}+(u-1)|-\frac{1}{\sqrt{2}} \ln |\sqrt{2}-(u-1)| \\ \\ \frac{1}{\sqrt{2}} \left( \ln |\sqrt{2}+u-1|-\ln |\sqrt{2}-u+1| \right)+C \\ \\ \frac{1}{\sqrt{2}} \left(\ln \left|\tan \frac{x}{2}+\sqrt{2}-1\right|-\ln \left|\sqrt{2}+1-\tan \frac{x}{2}\right|\right) +C \\ \\ \frac{1}{\sqrt{2}}\ln\left|\frac{\tan\frac{x}{2}+\sqrt{2}-1}{\sqrt{2}+1-\tan\frac{x}{2}}\right|+C \end{align} \]标签：right,frac,int,不定积分,sqrt,ln,例题,有理函数,left From： https://www.cnblogs.com/Preparing/p/18262755