Vtyjkyuk

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:

$$frac3dx3x=frac5dx5x$$
$$u=3x$$and$$v=5x$$
$$fracduu=fracdvv$$
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 $fracduu$, 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 $ln(frac 3x3)=ln(frac 5x 5)$, which simplifies to $ln(x)=ln(x)$.

To turn my comment into an answer, your proof is correct up to the step
$$ fracmathrmd uu = fracmathrmd vv $$

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

$$ ln u + mathrmC_1 = ln v + mathrmC_2 $$

Substituting $mathrmC = mathrmC_2 - mathrmC_1$ gives us:

$$beginalign
ln u &= ln v + mathrmC \
ln u - ln v &= mathrmC \
endalign$$

Substituting for $mathrmC$ 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 $mathrmC$, $ln u - ln v$, into $ln fracuv$ and then substitute that into the other equation above to get:

$$beginalign
ln u &= ln v + ln fracuv \
&= ln left( v cdot fracuv right)
endalign$$

Now take the exponential of both sides.

$$beginalign
u &= v cdot fracuv \
3x &= 5x cdot frac3x5x \
3 &= 5 cdot frac35
endalign$$

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.”