Answer
$$\int \frac{1}{(x+a)(x+b)}dx=\frac{1}{b-a}ln\left | \frac{x+a}{x+b} \right |+C$$
Work Step by Step
$$\int \frac{1}{(x+a)(x+b)}dx=\int \frac{1}{b-a}\frac{(x+b)-(x+a)}{(x+a)(x+b)}dx$$
$$=\frac{1}{b-a}\int (\frac{1}{x+a}-\frac{1}{x+b})dx$$
$$=\frac{1}{b-a}(ln\left | x+a \right |-ln\left | x+b \right |)+C$$
$$=\frac{1}{b-a}ln\left | \frac{x+a}{x+b} \right |+C$$
