A Difficult Integral

I'm wanting to keep up my LaTeX skills so I'm returning to an integral, specifically: $$\underset{0}{\overset{1}{\int }}\frac{x}{x^3+1}\text{d}x$$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:$$\frac{1}{3}\underset{0}{\overset{1}{\int }}\frac{x+1}{x^2-x+1}\text{d}x-\frac{1}{3}\underset{0}{\overset{1}{\int }}\frac{1}{x+1}\text{d}x$$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:$$\frac{1}{3}\underset{0}{\overset{1}{\int }}\frac{x+1}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}\text{d}x$$This newly changed denominator can then be divided into the two parts of the numerator to yield:$$\frac{1}{3}\underset{0}{\overset{1}{\int }}\frac{x}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}+\frac{1}{3}\underset{0}{\overset{1}{\int }}\frac{1}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}$$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: $$\frac{1}{6}(\ln |4x^2-4x+4|+2\sqrt{3}{\tan }^{-1}(\frac{2}{\sqrt{3}}(x-\frac{1}{2})))-\frac{1}{3}\ln |x+1|+C$$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.