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

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

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

Identifying coefficients:

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

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

On solving these simultaneous equations we obtain $\displaystyle \simplify[std]{A = {d-a*c}/{b-a}}$ and $\displaystyle \simplify[std]{B={d-b*c}/{a-b}}$

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

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