<s>[INST] Integrating 1/x 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.
 [/INST] 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.
</s>