Can somebody explain the solution to this indefinite integral

Question

I came across this problem while I was practicing for my math exam, and I could really use someone's help... The problem is:

$$\int\frac{1}{ax+b}\,dx$$

The solution by Symbolab and my professor is:

$$\frac{1}{a}ln|ax+b|$$

Can somebody explain why they took out the constant $\frac{1}{a}$?

Also, why is it $ax + b$ in the absolute value brackets in the solution, and not $x + \frac{b}{a}$

Thank you in advance!

Answer 1

Because $(\ln|x|)'=\frac{1}{x}$ and $$\left(\frac{1}{a}\ln|ax+b|\right)'=\frac{1}{a}\cdot\frac{1}{ax+b}\cdot (ax+b)'=\frac{1}{a}\cdot\frac{1}{ax+b}\cdot a=\frac{1}{ax+b}.$$

Answer 2

use that $$\frac{1}{ax+b}=\frac{1}{a}\left(\frac{1}{x+\frac{b}{a}}\right)$$

Answer 3

Note $$(\log f(x))' = \frac{f'(x)}{f(x)}.$$ Since $(ax+b)' = a$, to apply the formula above you need to write $$\int \frac{1}{ax+b} dx = \frac{1}{a} \int \frac{a}{ax+b} dx.$$