Integrating 1/x

Question

The standard definition of integrating $\frac{1}{x}$ is:

$$ \int \frac{dx}{ax + b} = \frac {1}{a} \ln |ax + b| + K $$

Now, if I'm understanding the "constant factor rule", that is: $$ \int k \frac{dy}{dx} dx = k\int \frac{dy}{dx} dx $$

Then what if we construct a new fraction:

$$ \frac{1}{ax + b} = \frac{1}{a(x + \frac{b}{a})} $$

Then we integrate, using the "constant factor rule", such that:

$$ \int \frac{dx}{a(x + \frac{b}{a})} = \frac {1}{a} \int \frac{dx}{(x + \frac{b}{a})} = \frac{1}{a} \ln |x+\frac{b}{a}| + K $$

Given that

$$ \frac{1}{a(x + \frac{b}{a})} =\frac{1}{ax + b} $$

Why is it that their integrals are not equal?

$$ \frac{1}{a} \ln |x+\frac{b}{a}| + K = \int \frac{dx}{a(x + \frac{b}{a})} \neq \int \frac{dx}{ax + b} = \frac {1}{a} \ln |ax + b| + K $$

I'm pretty sure I'm missing something super obvious here, I just don't know what.

Answer 1

But they are equal:

$$\ln|ax + b| = \ln\Big(|a| |x + \frac b a|\Big) = \ln |a| + \ln |x + \frac b a|$$

Now just roll $\ln |a|$ into the arbitrary constant.

Answer 2

Like T. Bongers said, they are equal. Your two constants, however, are not. When comparing different setups of the same integral, I highly suggest setting different names for your final constants so you won't forget that the first $K$ is actually the second $K$ plus $\ln |a|$.