$\dfrac{1}{b-a}\int \bigg ( \dfrac{1}{x+a} -\dfrac{1}{x+b} \bigg )dx = \dfrac{1}{b-a}\int \dfrac{1}{x+a}dx - \dfrac{1}{b-a}\int \dfrac{1}{x+a}dx $
$=\dfrac{1}{b-a}\int \dfrac{d(1+a)}{x+a} - \dfrac{1}{b-a}\int \dfrac{d(1+b)}{x+a} = \dfrac{1}{b-a} \ln |1+a|-\dfrac{1}{b-a} \ln |1+b|$
$=\dfrac{1}{b-a} \ln \bigg |\dfrac{1+a}{1+b} \bigg |+C$