Answer
$$\frac{3}{{2\left( {x + 1} \right)}} - \frac{3}{{2\left( {x + 3} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{3}{{\left( {x + 1} \right)\left( {x + 3} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{3}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{x + 3}} \cr
& {\text{Multiply each side by }}\left( {x + 1} \right)\left( {x + 3} \right) \cr
& 3 = A\left( {x + 3} \right) + B\left( {x + 1} \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Let }}x = - 1{\text{ into the equation }}\left( 1 \right) \cr
& 3 = A\left( { - 1 + 3} \right) + B\left( { - 1 + 1} \right) \cr
& 3 = 2A \cr
& A = \frac{3}{2} \cr
& {\text{Let }}x = - 3{\text{ into the equation }}\left( 1 \right) \cr
& 3 = A\left( { - 3 + 3} \right) + B\left( { - 3 + 1} \right) \cr
& 3 = - 2B \cr
& B = - \frac{3}{2} \cr
& \cr
& {\text{Use }}A{\text{ and }}B{\text{ to find the partial fraction decomposition}} \cr
& \frac{3}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \frac{{3/2}}{{x + 1}} + \frac{{ - 3/2}}{{x + 3}} \cr
& \frac{3}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \frac{3}{{2\left( {x + 1} \right)}} - \frac{3}{{2\left( {x + 3} \right)}} \cr} $$
