Answer
$$\int\frac{dx}{ax+b}=\frac{1}{a}\ln|ax+b|+c$$
Work Step by Step
To evaluate the integral $$\int\frac{dx}{ax+b}$$
we will use substitution $ax+b=t$ which gives us $adx=dt\Rightarrow dx=\frac{dt}{a}$ (we now that $a\neq0$). Putting this into the integral we get:
$$\int\frac{dx}{ax+b}=\int\frac{1}{t}\frac{dt}{a}=\frac{1}{a}\int\frac{1}{t}dt=\frac{1}{a}\ln|t|+c$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$:
$$\int\frac{dx}{ax+b}=\frac{1}{a}\ln|t|+c=\frac{1}{a}\ln|ax+b|+c$$
