Answer
$$\int\frac{dx}{5-3x}=-\frac{1}{3}\ln|5-3x|+c$$
Work Step by Step
To evaluate the integral $$\int\frac{dx}{5-3x}$$
we will use substitution $5-3x=t$ which gives us $-3dx=dt\Rightarrow dx=-\frac{dt}{3}$. Putting this into the integral we get:
$$\int\frac{dx}{5-3x}=\int\frac{1}{t}\Big(-\frac{dt}{3}\Big)=-\frac{1}{3}\int \frac{1}{t}dt=-\frac{1}{3}\ln|t|+c$$
where $c$ is arbitrary constant. Now we have to express our solution in terms of $x$:
$$\int\frac{dx}{5-3x}=-\frac{1}{3}\ln|t|+c=-\frac{1}{3}\ln|5-3x|+c$$
