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:$$ \int \! \Big ( \frac{\frac{1}{2} i}{x-i} - \frac{\frac{1}{2} i}{x+i} \Big ) \, dx$$We can now split the integral into two parts and take out the common factor:$$ \tfrac{1}{2} i \int \! \frac{1}{x-i} \, dx - \tfrac{1}{2} i \int \! \frac{1}{x+i} \, dx$$This is an easy integration, that gives the following:$$ \tfrac{1}{2} i \, (\ln(x-i)-\ln(x+i))$$This simplifies to:$$ \tfrac{1}{2} i \, \ln \Big ( \dfrac{x-i}{x+i} \Big ) \text{ . . .  key expression}$$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:$$ \begin{align} x+i &= \sqrt{x^2+1} \, e^{\, i \, \arctan(1/x)} \\x+i &= \sqrt{x^2+1} \, e^{\, -i \,\arctan(1/x)} \end{align}$$Substituting these expressions back into the earlier marked key expression gives:$$ \begin{align}  \tfrac{1}{2} \, i \, \ln \Big ( e^{ \,2i  \arctan(1/x) } \Big ) &= -\arctan \Big ( \dfrac{1}{x} \Big ) \\ &= \arctan(x) \pm \dfrac{\pi}{2} \end{align} $$So finally we have:$$ \begin{align} \int \! \frac{1}{1+x^2} \, dx &= \arctan(x) \pm \dfrac{\pi}{2} + C' \\ &= \arctan(x) + C \end{align}$$

A Difficult Integral

I'm wanting to keep up my LaTeX skills so I'm returning to an integral, specifically: $$  \int\limits_0^1 \, \frac{x}{x^3+1}\text{d}x$$The integral is not all that difficult really but it does involve a fair number of steps to get it right. Here is a graph of what is being attempted:

I used symbolab to work through the solution which turns out to be:\[\frac{1}{9}(\sqrt{3}\pi-\ln{(8)})\]The first step in the solution is to factorise the denominator. This is easy because \(x^3+1 \) is a sum of two cubes and so can be written as \((x+1)(x^2-x+1)\). Using partial fractions, the integral can be converted into the following form:\[ \frac{1}{3} \int\limits_0^1 \frac{x+1}{x^2-x+1} \text{d}x-\frac{1}{3} \int\limits_0^1 \frac{1}{x+1} \text{d}x \]The RHS integral is straightforward but the LHS one poses more difficulties. The way forward with the former is to write the quadratic in the denominator as a sum of two squares:\[ \frac{1}{3} \int\limits_0^1  \frac{x+1}{(x-\frac{1}{2})^2+\frac{3}{4}} \text{d}x \]This newly changed denominator can then be divided into the two parts of the numerator to yield:\[ \frac{1}{3} \int\limits_0^1 \frac{x}{(x-\frac{1}{2})^2+\frac{3}{4}}+\frac{1}{3} \int\limits_0^1 \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}} \]I'm not going to go into the details here but, with the appropriate substitutions, the LHS of this integral can be integrated to a logarithm and the RHS to an arctan using the formulae:\[ \int \frac{2t}{t^2+1} \text{d}t = \ln|t^2+1| + C \text{ and }\int \frac{1}{t^2+1} \text{d}t=\tan \!^{-1}{t}+C \]The indefinite integral for \( x/(x^3+1) \) turns out to be: \[ \frac{1}{6} \Big( \ln|4x^2-4x+4|+2\sqrt{3} \, \tan^{-1} \big( \frac{2}{\sqrt{3}}(x-\frac{1}{2}) \big) \Big)-\frac{1}{3}\ln|x+1|+C \]The definite integral for the limits of 0 and 1 is what was quoted earlier at the beginning of this post. In conclusion, I must say that Blogger is still inconsistently handling the display mode for LaTeX: sometimes it renders the LaTeX using the Javascript and other times it returns an image of the rendered code. Annoying.