$\displaystyle\int\frac{1}{x^3-1}dx$. I've had this in mind for days... I first write the integrand in partial fractions:$$\frac{1}{x^3-1}=\frac{A}{x-1}+\frac{B}{x-\omega}+\frac{C}{x-\omega^2}=\frac{(A+B+C)x^2+(A+B\omega+c\omega^2)x+(A+B\omega^2+C\omega)}{(x-1)(x-\omega)(x-\omega^2)}$$where $\omega$ is a primitive $3$rd root of unity. By comparing coefficients and then applying Gaussian elimination, I have $\{A=\frac{1}{3},B=\frac{\omega}{3},C=\frac{\omega^2}{3}$. Thus$$\displaystyle\int\frac{1}{x^3-1}dx=\frac{1}{3}\operatorname{log}(x-1)+\frac{\omega}{3}\operatorname{log}(x-\omega)+\frac{\omega^2}{3}\operatorname{log}(x-\omega^2)+C$$At first glance, I thought this method is brilliant. It is easy to be understood and generalised, but then I thought: Will I run into some problems? Because I know that complex logarithm has an apparent branch cut. Logarithm is not a well-defined function on $\Bbb C$ unless we specify the use of principal argument in its evaluation. Can somebody tell if this approach is appropriate or not?
P.S. Apparently I'm not familiar with topics in complex analysis, and I won't be able to learn such advanced things until after two or three years in university. :(