For $$x > 0$$, let $$f(x) = \int_1^x \frac{\log t}{1-t} dt$$. Then $$f(x) + f\left(\frac{1}{x}\right)$$ is equal to

We are given $$f(x) = \displaystyle\int_1^x \frac{\log t}{1 - t}\,dt$$ for $$x > 0$$, and we need to find $$f(x) + f\!\left(\dfrac{1}{x}\right)$$.

In $$f\!\left(\dfrac{1}{x}\right) = \displaystyle\int_1^{1/x} \frac{\log t}{1-t}\,dt$$, substitute $$t = \dfrac{1}{u}$$, so $$dt = -\dfrac{1}{u^2}\,du$$. When $$t = 1$$, $$u = 1$$; when $$t = \dfrac{1}{x}$$, $$u = x$$. Also $$\log t = -\log u$$ and $$1 - t = 1 - \dfrac{1}{u} = \dfrac{u-1}{u}$$.

Substituting: $$f\!\left(\frac{1}{x}\right) = \displaystyle\int_1^x \frac{-\log u}{\;\dfrac{u-1}{u}\;} \cdot \frac{1}{u^2}\,du = \int_1^x \frac{-\log u}{u(u-1)}\,du$$.

Using partial fractions, $$\dfrac{1}{u(u-1)} = \dfrac{-1}{u} + \dfrac{1}{u-1}$$, so $$\dfrac{-1}{u(u-1)} = \dfrac{1}{u} - \dfrac{1}{u-1} = \dfrac{1}{u} + \dfrac{1}{1-u}$$.

Therefore $$f\!\left(\frac{1}{x}\right) = \displaystyle\int_1^x \frac{\log u}{u}\,du + \int_1^x \frac{\log u}{1-u}\,du = \frac{(\log x)^2}{2} + f(x)$$.

Adding: $$f(x) + f\!\left(\frac{1}{x}\right) = f(x) + \frac{(\log x)^2}{2} + f(x) = 2f(x) + \frac{(\log x)^2}{2}$$.

To evaluate $$f(x)$$ in a useful form, note that $$\dfrac{\log t}{1-t} = -\log t \displaystyle\sum_{n=0}^{\infty} t^n$$ for $$|t| < 1$$. For $$0 < x < 1$$, this gives $$f(x) = -\displaystyle\sum_{n=0}^{\infty}\int_1^x t^n \log t\,dt$$. By integration by parts each term yields a series related to $$\displaystyle\sum \frac{x^{n+1}\log x}{n+1} - \frac{x^{n+1}-1}{(n+1)^2}$$. Through the standard dilogarithm identity $$\text{Li}_2(1-x) + \text{Li}_2(1-(1/x)) = -\dfrac{1}{2}(\log x)^2$$, and since $$f(x) = -\text{Li}_2(1-x)$$, we obtain $$f(x) + f(1/x) = \dfrac{1}{2}(\log x)^2$$.