The set of values of $m$ for which $mx^2-6mx+5m+1>0$ for all real $x$ is
$\textbf{(A)}~m<\frac14$
$\textbf{(B)}~m\ge0$
$\textbf{(C)}~0\le m\le\frac14$
$\textbf{(D)}~0\le m<\frac14$
To determine the set of values of \( m \) for which \( mx^2 - 6mx + 5m + 1 > 0 \) for all real \( x \), we need to analyze the quadratic expression. 

1. **Quadratic Form Analysis**:
   The given quadratic expression is \( mx^2 - 6mx + 5m + 1 \). For this quadratic to be positive for all real \( x \), the discriminant must be negative, and the leading coefficient \( m \) must be positive (since the parabola must open upwards).

2. **Discriminant Calculation**:
   The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c \) is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   For our quadratic equation \( mx^2 - 6mx + (5m + 1) \), we have:
   \[
   a = m, \quad b = -6m, \quad c = 5m + 1
   \]
   Therefore, the discriminant is:
   \[
   \Delta = (-6m)^2 - 4(m)(5m + 1) = 36m^2 - 4m(5m + 1)
   \]
   Simplifying the expression:
   \[
   \Delta = 36m^2 - 20m^2 - 4m = 16m^2 - 4m = 4m(4m - 1)
   \]

3. **Discriminant Condition**:
   For the quadratic to be positive for all \( x \), the discriminant must be negative:
   \[
   4m(4m - 1) < 0
   \]
   This inequality can be solved by analyzing the sign changes:
   \[
   4m(4m - 1) < 0 \implies m(4m - 1) < 0
   \]
   The critical points are \( m = 0 \) and \( m = \frac{1}{4} \). We test the intervals around these points:
   - For \( m < 0 \), \( m(4m - 1) \) is positive.
   - For \( 0 < m < \frac{1}{4} \), \( m(4m - 1) \) is negative.
   - For \( m > \frac{1}{4} \), \( m(4m - 1) \) is positive.

   Therefore, the inequality \( m(4m - 1) < 0 \) holds for:
   \[
   0 < m < \frac{1}{4}
   \]

4. **Boundary Conditions**:
   - At \( m = 0 \), the quadratic becomes \( 1 > 0 \), which is true.
   - At \( m = \frac{1}{4} \), the quadratic becomes \( \left(\frac{1}{4}\right)x^2 - \frac{3}{2}x + \frac{5}{4} + 1 = \frac{1}{4}(x^2 - 6x + 6) \). The discriminant of \( x^2 - 6x + 6 \) is \( 36 - 24 = 12 \), which is positive, so it has real roots and can be zero.

   Therefore, \( m = \frac{1}{4} \) does not satisfy the condition of being positive for all \( x \).

Conclusion:
The set of values of \( m \) for which \( mx^2 - 6mx + 5m + 1 > 0 \) for all real \( x \) is:
\[
0 \leq m < \frac{1}{4}
\]

The final answer is \(\boxed{0 \leq m < \frac{1}{4}}\).