An indefinite integral with no constant of integration

[Mark44]

What you have is a differential equation. To find a unique solution of a first-order differential equation, you need an initial condition, say m(a) = m_a.

Before continuing I should add that m(x) really represents the mass density of your rod. The actual mass of a very short section of length Δx would be m(x)Δx.

Your problem, with the addition of an initial condition, is
$$\frac{m'}{m} = \frac{1}{x}, m(a) = m_a$$

From this we get ln(m) = ln(x) + C. Exponentiating both sides, we get m = e^{ln(x) + C} = e^C e^ln(x) = Kx, where K = e^C.

So m(x) = Kx.

From the initial condition, m(a) = m_a = Ka $\Rightarrow$ K = m_a/a.

The unambiguous solution is m(x) = m_a/a * x. You can verify that this function satisfies the differential equation and initial condition.

To find the mass of the rod in the interval [a, b], calculate integral of m(x) between a and b.