Answer
$$3\ln \left| {x + 1} \right| + \ln \left| {2x + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{8x + 5}}{{2{x^2} + 3x + 1}}} dx \cr
& = \int {\frac{{8x + 5}}{{\left( {x + 1} \right)\left( {2x + 1} \right)}}} dx \cr
& {\text{partial fractions}} \cr
& \frac{{8x + 5}}{{\left( {x + 1} \right)\left( {2x + 1} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{2x + 1}} \cr
& {\text{multiplying}} \cr
& 8x + 5 = A\left( {2x + 1} \right) + B\left( {x + 1} \right) \cr
& x = - 1 \to A = 3 \cr
& x = - \frac{1}{2} \to B = 2 \cr
& {\text{substituting constants}} \cr
& \frac{A}{{x + 1}} + \frac{B}{{2x + 1}} = \frac{3}{{x + 1}} + \frac{2}{{2x + 1}} \cr
& \int {\frac{{8x + 5}}{{2{x^2} + 3x + 1}}} dx = \int {\left( {\frac{3}{{x + 1}} + \frac{2}{{2x + 1}}} \right)dx} \cr
& {\text{integrating}} \cr
& = 3\ln \left| {x + 1} \right| + \ln \left| {2x + 1} \right| + C \cr} $$
