Answer
$$ - \frac{{2 + 3x}}{{4x\left( {4 + 3x} \right)}} - \frac{3}{{16}}\ln \left| {\frac{x}{{4 + 3x}}} \right| + C{\text{ }}$$
Work Step by Step
$$\eqalign{
& \int {\frac{2}{{{x^2}{{\left( {4 + 3x} \right)}^2}}}} dx \cr
& {\text{Use a table of integrals with forms involving }}a + bu \cr
& {\text{From the table of integrals in the back of the book}} \cr
& {\text{Formula 13}} \cr
& \int {\frac{1}{{{u^2}{{\left( {a + bu} \right)}^2}}}du = - \frac{1}{{{a^2}}}\left[ {\frac{{a + 2bu}}{{u\left( {a + bu} \right)}} + \frac{{2b}}{a}\ln \left| {\frac{u}{{a + bu}}} \right|} \right] + C} \cr
& \int {\frac{2}{{{x^2}{{\left( {4 + 3x} \right)}^2}}}} dx \to a = 4,{\text{ }}b = 3,{\text{ }}u = x,{\text{ }}du = dx \cr
& {\text{Substituting into formula 13}} \cr
& \int {\frac{2}{{{x^2}{{\left( {4 + 3x} \right)}^2}}}} dx = - \frac{2}{{{{\left( 4 \right)}^2}}}\left[ {\frac{{4 + 2\left( 3 \right)\left( x \right)}}{{x\left( {4 + 3x} \right)}} + \frac{{2\left( 3 \right)}}{4}\ln \left| {\frac{x}{{4 + 3x}}} \right|} \right] + C \cr
& {\text{Multiply and simplify}} \cr
& = - \frac{1}{8}\left[ {\frac{{4 + 6x}}{{x\left( {4 + 3x} \right)}} + \frac{3}{2}\ln \left| {\frac{x}{{4 + 3x}}} \right|} \right] + C \cr
& = - \frac{{4 + 6x}}{{8x\left( {4 + 3x} \right)}} - \frac{3}{{16}}\ln \left| {\frac{x}{{4 + 3x}}} \right| + C \cr
& = - \frac{{2 + 3x}}{{4x\left( {4 + 3x} \right)}} - \frac{3}{{16}}\ln \left| {\frac{x}{{4 + 3x}}} \right| + C{\text{ }} \cr} $$
