A Beautiful Integral

We know that for $|x|<1,$ $\displaystyle\sum_{n=0}^\infty x^n=\dfrac{1}{1-x}.$ Integrating this, we find that $\ln(1-x)=-\displaystyle\sum_{n=1}\dfrac{x^n}{n}.$ Note that the constant of integration has been omitted because when $x=0,$ $\ln(1-x)=-\displaystyle\sum_{n=1}\dfrac{x^n}{n}=0.$ Thus, we can say the following: $\frac{\ln(1-x)}{x}=-\sum_{n=1}^\infty\frac{x^{n-1}}{n}$ Finally, we can evaluate the integral. $\int_0^1\frac{\ln(1-x)}{x}\text{ }dx=-\int_0^1\sum_{n=1}^\infty\frac{x^{n-1}}{n}\text{ }dx=\left.\sum_{n=1}^\infty\dfrac{x^n}{n^2}\right|^1_0=-\sum_{n=1}^\infty\dfrac{1}{n^2}$ This is a familiar sum equal to $-\dfrac{\pi^2}{6}.$ Thus, $\{A,B,C\}=\{1,6,2\}$ and $A+B+C=\boxed{9}$ .

The given integral is well known which is:

$\large{Li_{2}(1)=\frac{\pi^{2}}{6}}$

where $Li_{2}(x)$ is dilogarithm function

Expanding ln(1-x) as -(x+x^2/2 +x^3/3....) Dividing by x and inegrating each of expressions within limits we get final expressions as -(1+1/4 +1+1/9... )Which is nothing but- pi^2/6. Hence A+B +C =9