I am reading Maxwell's treatise Article 524.
There, he equates two equations (31) and (36):
$$\oint^{s'}_0 \oint^{s}_0 \rho \ {\vec{ds}.\vec{ds'}}=\oint^{s'}_0 \oint^{s}_0 \dfrac{\vec{ds}.\vec{ds'}}{r}$$
Then he says:
$$\rho=\dfrac{1}{r}$$
How can this be? For example, in the following equation, the definite integrals, the variables and limits of integration are same. But the integrand is different. $$\int_0^\pi \sin x\,dx = 2 = \int_0^\pi \frac 2\pi\, dx$$
So how can $\rho=\dfrac{1}{r}?$
Not the most rigorous, but I hope the following sketch proof gives you an idea of why it should be true.
Suppose you have a sufficiently "nice" domain $D$ and sufficiently "well-behaved" functions $f$ and $g$ defined on $D$ such that
$$ \int _C fds=\int_Cgds$$
for all sufficiently "nice" $C \subset D$, where $ds$ is a differential element of $D$. Then certainly,
$$\int_C (f-g)ds=0 \; \; \; \; \forall C \subset D$$
Assume for contradiction that there exists $x_0 \in D$ such that $(f-g)(x_0) \neq 0$.
Suppose that $(f-g)(x_0)>0$. Since $f-g$ is sufficiently "well-behaved" (here we need that it is continuous), it follows that there exists a small region $S \subset D$ around $x_0$ such that $(f-g)(x)>0$ for all $x \in S$, so that
$$\int_S (f-g)ds>0$$
which gives a contradiction. We get a similar contradiction if $(f-g)(x_0)<0$.
Thus, by contradiction, it follows that $(f-g)(x)=0$ for all $x \in D$, hence $f=g$ in $D$ as required.