1. The problem is to evaluate the definite integral $$\int_0^1 x \ln(x+1) \, dx.$$\n\n2. Use integration by parts. Let \(u = \ln(x+1)\) and \(dv = x \, dx\). Then \(du = \frac{1}{x+1} \, dx\) and \(v = \frac{x^2}{2}\).\n\n3. Applying integration by parts formula: $$\int u \, dv = uv - \int v \, du,$$ we get\n$$\int_0^1 x \ln(x+1) \, dx = \left. \frac{x^2}{2} \ln(x+1) \right|_0^1 - \int_0^1 \frac{x^2}{2} \cdot \frac{1}{x+1} \, dx.$$\n\n4. Evaluate the boundary term: $$\left. \frac{x^2}{2} \ln(x+1) \right|_0^1 = \frac{1}{2} \ln 2 - 0 = \frac{\ln 2}{2}.$$\n\n5. Simplify the integral: $$\int_0^1 \frac{x^2}{2(x+1)} \, dx = \frac{1}{2} \int_0^1 \frac{x^2}{x+1} \, dx.$$\n\n6. Perform polynomial division on \(\frac{x^2}{x+1}\):\n$$\frac{x^2}{x+1} = x - 1 + \frac{1}{x+1}.$$\n\n7. Substitute back: $$\frac{1}{2} \int_0^1 (x - 1 + \frac{1}{x+1}) \, dx = \frac{1}{2} \left( \int_0^1 x \, dx - \int_0^1 1 \, dx + \int_0^1 \frac{1}{x+1} \, dx \right).$$\n\n8. Evaluate each integral:\n- $$\int_0^1 x \, dx = \frac{1}{2}.$$\n- $$\int_0^1 1 \, dx = 1.$$\n- $$\int_0^1 \frac{1}{x+1} \, dx = \ln 2.$$\n\n9. Substitute these values: $$\frac{1}{2} \left( \frac{1}{2} - 1 + \ln 2 \right) = \frac{1}{2} \left( -\frac{1}{2} + \ln 2 \right) = \frac{\ln 2}{2} - \frac{1}{4}.$$\n\n10. Combine all parts: $$\int_0^1 x \ln(x+1) \, dx = \frac{\ln 2}{2} - \left( \frac{\ln 2}{2} - \frac{1}{4} \right) = \frac{1}{4}.$$\n\nFinal answer: $$\boxed{\frac{1}{4}}.$$