Answer
$$\ln \left| {x - 3} \right| + \frac{2}{3}\ln \left| {3x + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{5x - 5}}{{3{x^2} - 8x - 3}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{5x - 5}}{{3{x^2} - 8x - 3}} = \frac{{5x - 5}}{{\left( {x - 3} \right)\left( {3x + 1} \right)}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{5x - 5}}{{\left( {x - 3} \right)\left( {3x + 1} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{3x + 1}} \cr
& {\text{Multiplying the equation by }}\left( {x - 3} \right)\left( {3x + 1} \right){\text{, we have}} \cr
& 5x - 5 = A\left( {3x + 1} \right) + B\left( {x - 3} \right) \cr
& {\text{if we set }}x = 3 \cr
& 5\left( 3 \right) - 5 = A\left( {3\left( 3 \right) + 1} \right) + B\left( 0 \right) \cr
& 10 = A\left( {10} \right) \cr
& A = 1 \cr
& {\text{if we set }}x = - 1/3 \cr
& 5\left( { - 1/3} \right) - 5 = A\left( 0 \right) + B\left( { - \frac{1}{3} - 3} \right) \cr
& B = 2 \cr
& {\text{then}} \cr
& \frac{{5x - 5}}{{\left( {x - 3} \right)\left( {3x + 1} \right)}} = \frac{1}{{x - 3}} + \frac{2}{{3x + 1}} \cr
& \cr
& \int {\frac{{5x - 5}}{{3{x^2} - 8x - 3}}} dx = \int {\left( {\frac{1}{{x - 3}} + \frac{2}{{3x + 1}}} \right)dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\frac{1}{{x - 3}}dx} + \frac{2}{3}\int {\frac{3}{{3x + 1}}} dx \cr
& {\text{integrating}} \cr
& = \ln \left| {x - 3} \right| + \frac{2}{3}\ln \left| {3x + 1} \right| + C \cr} $$
