To evaluate the given triple integral:

$\int_0^1 \int_0^x \int_y^x (3x + 5y) \, dz \, dy \, dx$

we can proceed by evaluating the innermost integral with respect to $z$, then $y$, and finally $x$.

1. Integrate with respect to $z$:

   $\int_y^x (3x + 5y) \, dz = (3x + 5y) \int_y^x 1 \, dz = (3x + 5y) [z]_y^x$

   Substituting the limits for $z$:

   $= (3x + 5y) (x - y) = 3x(x - y) + 5y(x - y)$

   Expanding this:

   $= 3x^2 - 3xy + 5yx - 5y^2 = 3x^2 + 2xy - 5y^2$

   So now the integral becomes:

   $\int_0^1 \int_0^x (3x^2 + 2xy - 5y^2) \, dy \, dx$

2. Integrate with respect to $y$:

   We now need to evaluate:

   $\int_0^1 \int_0^x (3x^2 + 2xy - 5y^2) \, dy \, dx$

   Split it term by term:

   $= \int_0^1 \int_0^x 3x^2 \, dy \, dx + \int_0^1 \int_0^x 2xy \, dy \, dx - \int_0^1 \int_0^x 5y^2 \, dy \, dx$

   Evaluating each part:

   • For $\int_0^x 3x^2 \, dy = 3x^2 y \Big|_0^x = 3x^3$
   • For $\int_0^x 2xy \, dy = 2x \cdot \frac{y^2}{2} \Big|_0^x = x \cdot x^2 = x^3$
   • For $\int_0^x 5y^2 \, dy = 5 \cdot \frac{y^3}{3} \Big|_0^x = \frac{5x^3}{3}$

   Combining these, we have:

   [ \int_0^1 (3x^3 + x^3 - \frac{5x^3}{3}) , dx = \int_0^1 \frac{9x^3}{3} , dx = \int_0^1