1. **State the problem:** Evaluate the triple integral $$\int_0^1 \int_0^1 \int_0^{x+y} \ln z \, dz \, dy \, dx.$$\n\n2. **Integrate with respect to $z$: ** We first compute $$\int_0^{x+y} \ln z \, dz.$$\nRecall the integral formula: $$\int \ln z \, dz = z \ln z - z + C.$$\nApplying the limits, we get:\n$$\int_0^{x+y} \ln z \, dz = \left[(x+y) \ln(x+y) - (x+y)\right] - \lim_{z \to 0^+} (z \ln z - z).$$\nAs $z \to 0^+$, $z \ln z \to 0$ and $z \to 0$, so the limit is 0.\nThus, the integral simplifies to:\n$$ (x+y) \ln(x+y) - (x+y).$$\n\n3. **Rewrite the integral:** Now the triple integral reduces to a double integral:\n$$\int_0^1 \int_0^1 \left[(x+y) \ln(x+y) - (x+y)\right] dy \, dx.$$\n\n4. **Split the integral:**\n$$\int_0^1 \int_0^1 (x+y) \ln(x+y) \, dy \, dx - \int_0^1 \int_0^1 (x+y) \, dy \, dx.$$\n\n5. **Evaluate the simpler integral:**\n$$\int_0^1 \int_0^1 (x+y) \, dy \, dx = \int_0^1 \left[ x y + \frac{y^2}{2} \right]_0^1 dx = \int_0^1 \left(x + \frac{1}{2}\right) dx = \left[ \frac{x^2}{2} + \frac{x}{2} \right]_0^1 = \frac{1}{2} + \frac{1}{2} = 1.$$\n\n6. **Focus on the integral:** $$I = \int_0^1 \int_0^1 (x+y) \ln(x+y) \, dy \, dx.$$\n\n7. **Change variables inside the integral:** Let $t = x + y$. For fixed $x$, as $y$ goes from 0 to 1, $t$ goes from $x$ to $x+1$. Then $dy = dt$. So\n$$\int_0^1 (x+y) \ln(x+y) \, dy = \int_x^{x+1} t \ln t \, dt.$$\n\n8. **Rewrite $I$ as:**\n$$I = \int_0^1 \int_x^{x+1} t \ln t \, dt \, dx.$$\n\n9. **Change the order of integration:** The region for $(x,t)$ is $0 \le x \le 1$, $x \le t \le x+1$.\nFor $t$, the minimum is when $x=0$, $t \ge 0$, maximum when $x=1$, $t \le 2$.\nFor each $t$, $x$ varies from $0$ to $t$ if $t \le 1$, and from $t-1$ to $1$ if $t > 1$.\n\nSo split $I$ as:\n$$I = \int_0^1 \int_0^t t \ln t \, dx \, dt + \int_1^2 \int_{t-1}^1 t \ln t \, dx \, dt.$$\n\n10. **Integrate with respect to $x$: **\nFor $0 \le t \le 1$, $x$ goes from 0 to $t$, so length is $t$.\nFor $1 < t \le 2$, $x$ goes from $t-1$ to 1, length is $1 - (t-1) = 2 - t$.\n\nThus,\n$$I = \int_0^1 t \ln t \cdot t \, dt + \int_1^2 t \ln t \cdot (2 - t) \, dt = \int_0^1 t^2 \ln t \, dt + \int_1^2 t (2 - t) \ln t \, dt.$$\n\n11. **Evaluate the first integral:**\n$$J_1 = \int_0^1 t^2 \ln t \, dt.$$\nUse integration by parts: Let $u = \ln t$, $dv = t^2 dt$, then $du = \frac{1}{t} dt$, $v = \frac{t^3}{3}$.\n\n$$J_1 = \left. \frac{t^3}{3} \ln t \right|_0^1 - \int_0^1 \frac{t^3}{3} \cdot \frac{1}{t} dt = 0 - \frac{1}{3} \int_0^1 t^2 dt = -\frac{1}{3} \cdot \frac{1}{3} = -\frac{1}{9}.$$\n\n12. **Evaluate the second integral:**\n$$J_2 = \int_1^2 t (2 - t) \ln t \, dt = \int_1^2 (2t - t^2) \ln t \, dt = 2 \int_1^2 t \ln t \, dt - \int_1^2 t^2 \ln t \, dt.$$\n\n13. **Evaluate $\int_1^2 t \ln t \, dt$: **\nUse integration by parts: $u = \ln t$, $dv = t dt$, $du = \frac{1}{t} dt$, $v = \frac{t^2}{2}$.\n\n$$\int_1^2 t \ln t \, dt = \left. \frac{t^2}{2} \ln t \right|_1^2 - \int_1^2 \frac{t^2}{2} \cdot \frac{1}{t} dt = \frac{4}{2} \ln 2 - 0 - \frac{1}{2} \int_1^2 t dt = 2 \ln 2 - \frac{1}{2} \cdot \frac{3}{2} = 2 \ln 2 - \frac{3}{4}.$$\n\n14. **Evaluate $\int_1^2 t^2 \ln t \, dt$: **\nUse integration by parts: $u = \ln t$, $dv = t^2 dt$, $du = \frac{1}{t} dt$, $v = \frac{t^3}{3}$.\n\n$$\int_1^2 t^2 \ln t \, dt = \left. \frac{t^3}{3} \ln t \right|_1^2 - \int_1^2 \frac{t^3}{3} \cdot \frac{1}{t} dt = \frac{8}{3} \ln 2 - 0 - \frac{1}{3} \int_1^2 t^2 dt = \frac{8}{3} \ln 2 - \frac{1}{3} \cdot \frac{7}{3} = \frac{8}{3} \ln 2 - \frac{7}{9}.$$\n\n15. **Substitute back into $J_2$: **\n$$J_2 = 2 \left(2 \ln 2 - \frac{3}{4}\right) - \left(\frac{8}{3} \ln 2 - \frac{7}{9}\right) = 4 \ln 2 - \frac{3}{2} - \frac{8}{3} \ln 2 + \frac{7}{9} = \left(4 - \frac{8}{3}\right) \ln 2 + \left(-\frac{3}{2} + \frac{7}{9}\right).$$\n\nCalculate coefficients:\n$$4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}, \quad -\frac{3}{2} + \frac{7}{9} = -\frac{27}{18} + \frac{14}{18} = -\frac{13}{18}.$$\n\nSo,\n$$J_2 = \frac{4}{3} \ln 2 - \frac{13}{18}.$$\n\n16. **Sum $I = J_1 + J_2$: **\n$$I = -\frac{1}{9} + \frac{4}{3} \ln 2 - \frac{13}{18} = \frac{4}{3} \ln 2 - \left(\frac{1}{9} + \frac{13}{18}\right) = \frac{4}{3} \ln 2 - \frac{15}{18} = \frac{4}{3} \ln 2 - \frac{5}{6}.$$\n\n17. **Recall the original integral:**\n$$\int_0^1 \int_0^1 \int_0^{x+y} \ln z \, dz \, dy \, dx = I - 1 = \left(\frac{4}{3} \ln 2 - \frac{5}{6}\right) - 1 = \frac{4}{3} \ln 2 - \frac{11}{6}.$$\n\n**Final answer:**\n$$\boxed{\frac{4}{3} \ln 2 - \frac{11}{6}}.$$