Answer
$$\ln 3$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\frac{3}{{4{x^2} + 5x + 1}}} dx \cr
& {\text{Integrating }}\int {\frac{3}{{4{x^2} + 5x + 1}}} dx \cr
& \frac{3}{{4{x^2} + 5x + 1}} = \frac{3}{{\left( {x + 1} \right)\left( {4x + 1} \right)}} \cr
& \frac{3}{{\left( {x + 1} \right)\left( {2x + 1} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{4x + 1}} \cr
& 3 = A\left( {4x + 1} \right) + B\left( {x + 1} \right) \cr
& 3 = 4Ax + A + Bx + B \cr
& 4A + B = 0 \cr
& A + B = 3 \cr
& {\text{Solving the system of equations simultaneously}} \cr
& A = - 1,{\text{ }}B = 4 \cr
& \frac{3}{{\left( {x + 1} \right)\left( {2x + 1} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{4x + 1}} \cr
& \frac{3}{{\left( {x + 1} \right)\left( {2x + 1} \right)}} = \frac{{ - 1}}{{x + 1}} + \frac{4}{{4x + 1}} \cr
& \int_0^2 {\frac{3}{{4{x^2} + 5x + 1}}} dx = \int_0^2 {\left( { - \frac{1}{{x + 1}} + \frac{4}{{4x + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ { - \ln \left| {x + 1} \right| + \ln \left| {4x + 1} \right|} \right]_0^2 \cr
& = \left[ {\ln \left| {\frac{{4x + 1}}{{x + 1}}} \right|} \right]_0^2 \cr
& = \ln \left| {\frac{{4\left( 2 \right) + 1}}{{2 + 1}}} \right| - \ln \left| {\frac{{4\left( 0 \right) + 1}}{{0 + 1}}} \right| \cr
& = \ln \left( 3 \right) - \ln \left( 1 \right) \cr
& = \ln 3 \cr} $$
