Where did I go wrong with my odd proof that $\frac{3dx}{3x} = \frac{5dx}{5x} \iff 3=5$?

I don't know where I went wrong, but it's interesting for me. Please check where my fault is! It is obvious that the below equation is correct:

$$\frac{3dx}{3x}=\frac{5dx}{5x}$$ $$u=3x$$and$$v=5x$$ $$\frac{du}{u}=\frac{dv}{v}$$ integrate both sides: $$ln(u)=ln(v)$$ $$u=v$$ $$3x=5x$$ so, $$3=5$$

Integrating we obtain

$$\ln(u)=\ln(v)+ C$$

If you're taking an indefinite integral, then you need to include constants. Integrating, we get $\ln(u)+C_1 = ln(v)+C_2$. We can collect $C_1$ and $C_2$ into one constant by setting $C_3 = C_2-C_1$, getting $\ln(u) = ln(v)+C_3$. We can then exponentiate both sides, getting $u=ve^{C_3}$. Setting $C_4=e^{C_3}$, this becomes $u=C_4v$. In this case, $C_4$ is $\frac35$.

If you're taking definite integrals, you need to include the limits. Suppose we integrate starting from $x = 1$. Then $u$ starts at $3$. So when we integrate $\frac{du}u$, we have $\ln(u)$ at the upper limit, but we have to subtract $\ln$ of the lower limit, and the lower limit is $3$. So we have that the definite integral is $\ln(u)-\ln(3)=\ln(\frac u3)$. Similarly, for $v$, we get that the definite integral is $\ln(\frac v 5)$. Plugging in $u=3x$ and $v=5x$, we get $\ln(\frac {3x}3)=\ln(\frac {5x} 5)$, which simplifies to $\ln(x)=\ln(x)$.

To turn my comment into an answer, your proof is correct up to the step $$ \frac{\mathrm{d} u}{u} = \frac{\mathrm{d} v}{v} $$

At that point, as others have said, you forget to add the constants of integration when you integrate:

$$ \ln u + \mathrm{C_1} = \ln v + \mathrm{C_2} $$

Substituting $\mathrm{C} = \mathrm{C}_2 - \mathrm{C}_1$ gives us:

$$\begin{align} \ln u &= \ln v + \mathrm{C} \\ \ln u - \ln v &= \mathrm{C} \\ \end{align}$$

Substituting for $\mathrm{C}$ in the first equation just gets us $\ln u = \ln u$, which is clearly no contradiction. A little more interesting, perhaps, is to turn our identity for $\mathrm{C}$, $\ln u - \ln v$, into $\ln \frac{u}{v}$ and then substitute that into the other equation above to get:

$$\begin{align} \ln u &= \ln v + \ln \frac{u}{v} \\ &= \ln \left( v \cdot \frac{u}{v} \right) \end{align}$$

Now take the exponential of both sides.

$$\begin{align} u &= v \cdot \frac{u}{v} \\ 3x &= 5x \cdot \frac{3x}{5x} \\ 3 &= 5 \cdot \frac{3}{5} \end{align}$$

This is presumably what Serge Seredenko meant when he said in his comment that 3 is equal to 5, “[i]n this case multiplied by a constant.”