Answer
$$\int\frac{dx}{ax+b}(a\ne0)=\frac{\ln|ax+b|}{a}+C$$
Work Step by Step
$$A=\int\frac{dx}{ax+b}(a\ne0)$$
Let $u=ax+b$.
We would have $du=(ax+b)'dx=adx$. Therefore, $dx=\frac{1}{a}du$
Substitute into $A$, we have $$A=\frac{1}{a}\int\frac{1}{u}du$$ $$A=\frac{1}{a}\ln|u|+C$$ $$A=\frac{\ln|ax+b|}{a}+C$$