1. **State the problem:** Determine whether the integral $$\int_0^1 \frac{x}{x-1} \, dx$$ converges or diverges. 2. **Analyze the integrand:** The integrand is $$f(x) = \frac{x}{x-1}$$. 3. **Check for discontinuities in the interval [0,1]:** The denominator $$x-1=0$$ at $$x=1$$, which is the upper limit of integration. Hence, there is a potential improper integral due to a vertical asymptote at the upper bound. 4. **Rewrite the integral as an improper integral:** $$ \int_0^1 \frac{x}{x-1} \, dx = \lim_{t \to 1^-} \int_0^t \frac{x}{x-1} \, dx $$ 5. **Simplify the integrand by algebraic manipulation:** $$ \frac{x}{x-1} = \frac{x-1+1}{x-1} = \frac{x-1}{x-1} + \frac{1}{x-1} = 1 + \frac{1}{x-1} $$ 6. **Now integrate term by term:** $$ \int_0^t \left(1 + \frac{1}{x-1}\right) dx = \int_0^t 1 \, dx + \int_0^t \frac{1}{x-1} \, dx = (t - 0) + \left[\ln|x-1|\right]_0^t = t + \ln|t-1| - \ln|0-1| $$ 7. **Evaluate $$\ln|0-1| = \ln 1 = 0$$ and rewrite:** $$ \int_0^t \frac{x}{x-1} \, dx = t + \ln|t-1| $$ 8. **Take the limit:** $$ \lim_{t \to 1^-} \left(t + \ln|t-1|\right) $$ As $$t \to 1^-$$, $$t \to 1$$ and $$t-1 \to 0^-$$. 9. **Behavior of the logarithm term:** $$\ln|t-1| = \ln(1-t)\quad \text{where }1-t \to 0^+,$$ and $$\ln(0^+) = -\infty$$. 10. **Hence, this limit is:** $$ 1 + (-\infty) = -\infty $$ 11. **Conclusion:** Since the integral tends to negative infinity, the improper integral diverges. **Final answer:** The integral $$\int_0^1 \frac{x}{x-1} \, dx$$ diverges because the integrand has a vertical asymptote at $$x=1$$ and the integral does not approach a finite limit there.