Using partial fractions we have to find $A$ and $B$ such that:
$$\simplify[std]{{c}/((x +{a})*(x+{b})) = A/(x+{a})+B/(x+{b})}$$
Multiplying both sides of the equation by $\displaystyle \simplify[std]{((x +{a})*(x+{b}))}$ we obtain:

$\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}}\Rightarrow \simplify[std]{(A+B)*x+{b}A+{a}B={c}}$. 

One method to find A and B is by comparing coefficients:

Identifying coefficients:

Constant term: $\simplify[std]{{b}*A+{a}*B = {c}}$

Coefficient of $x$: $\simplify[std]{A + B = 0}$ which gives $A = -B$

Solving these simultaneous equations gives $\displaystyle \simplify[std]{A = {c}/{b-a}}$ and $\displaystyle \simplify[std]{B={-c}/{b-a}}$

Which gives: $$\simplify[std]{{c}/((x +{a})*(x+{b})) = ({c}/{b-a})*(1/(x+{a}) -1/(x+{b}))}$$

So $$\begin{eqnarray*} I &=& \simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}\\ &=& \simplify[std]{({c}/{b-a})*(Int(1/(x+{a}),x) -Int(1/(x+{b}),x))}\\ &=& \simplify[std]{({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C} \end{eqnarray*}$$

Or equivalently:  $\displaystyle I= \simplify[std]{({c}/{b-a})*(ln((x+{a})/(x+{b})))+C}$