Answer
$$2{x^3} + {x^2} + \frac{1}{{2\left( {3x - 1} \right)}} + \frac{2}{{7\left( {x + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{6{x^5} + 7{x^4} - {x^2} + 2x}}{{3{x^2} + 2x - 1}} \cr
& {\text{Using the long division}} \cr
& \frac{{6{x^5} + 7{x^4} - {x^2} + 2x}}{{3{x^2} + 2x - 1}} = 2{x^3} + {x^2} + \frac{{2x}}{{3{x^2} + 2x - 1}} \cr
& {\text{Decompose }}\frac{{2x}}{{3{x^2} + 2x - 1}}{\text{ into partial fractions}} \cr
& \frac{{2x}}{{3{x^2} + 2x - 1}} = \frac{{2x}}{{\left( {3x - 1} \right)\left( {x + 1} \right)}} \cr
& \frac{{2x}}{{\left( {3x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{3x - 1}} + \frac{B}{{x + 1}} \cr
& 2x = A\left( {x + 1} \right) + B\left( {3x - 1} \right) \cr
& {\text{If }}x = 1/3 \cr
& \frac{2}{3} = A\left( {\frac{4}{3}} \right) \cr
& A = \frac{1}{2} \cr
& {\text{If }}x = - 1 \cr
& - 2 = B\left( { - 7} \right) \cr
& B = \frac{2}{7} \cr
& {\text{Use }}A,{\text{ and }}B{\text{ to find the partial fraction decomposition}} \cr
& \frac{{2x}}{{\left( {3x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{2\left( {3x - 1} \right)}} + \frac{2}{{7\left( {x + 1} \right)}} \cr
& \frac{{6{x^5} + 7{x^4} - {x^2} + 2x}}{{3{x^2} + 2x - 1}} = 2{x^3} + {x^2} + \frac{1}{{2\left( {3x - 1} \right)}} + \frac{2}{{7\left( {x + 1} \right)}} \cr} $$
