Solving a Well-Known Integral Without Trigometric Substitution

This approach to the integration of a well-known integral is explained in William Keen's article in Cantor's Paradise. I'm just following the steps that the article outlined and getting an opportunity to reinforce my LaTeX skills. Let's start with the integral:$$\int \frac{1}{1+x^2}dx$$Firstly, we factorise it, making use of complex numbers. This gives:$$\int \frac{1}{(x+i)(x-i)}dx$$Next we create partial fractions:$$\int (\frac{\frac{1}{2}i}{x-i}-\frac{\frac{1}{2}i}{x+i})dx$$We can now split the integral into two parts and take out the common factor:$$\frac{1}{2}i\int \frac{1}{x-i}dx-\frac{1}{2}i\int \frac{1}{x+i}dx$$This is an easy integration, that gives the following:$$\frac{1}{2}i(\ln (x-i)-\ln (x+i))$$This simplifies to:$$\frac{1}{2}i\ln ({\displaystyle \frac{x-i}{x+i}})\text{ . . . key expression}$$Figure 1 shows a diagram from the article that illustrates how the complex numbers $x+i$ can be changed into polar coordinates and, by extension, $x-i$.

Figure 1

Converting our complex numbers to polar coordinates, we find that:$$\begin{array}{rl}x+i& =\sqrt{x^2+1}{e}^{i\arctan (1/x)}\\ x+i& =\sqrt{x^2+1}{e}^{-i\arctan (1/x)}\end{array}$$Substituting these expressions back into the earlier marked key expression gives:$$\begin{array}{rl}\frac{1}{2}i\ln (e^{2i\arctan (1/x)})& =-\arctan ({\displaystyle \frac{1}{x}})\\ & =\arctan (x)\pm {\displaystyle \frac{\pi }{2}}\end{array}$$So finally we have:$$\begin{array}{rl}\int \frac{1}{1+x^2}dx& =\arctan (x)\pm {\displaystyle \frac{\pi }{2}}+C^{\prime }\\ & =\arctan (x)+C\end{array}$$