Answer
$$\ln \left( {\frac{3}{4}} \right) + 1$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} dv \cr
& {\text{By the definition of the improper integrals, we have }} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr
& {\text{then}}{\text{,}} \cr
& \int_1^\infty {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} dv = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} \cr
& \cr
& {\text{Integrate }}\int {\frac{{3v - 1}}{{4{v^3} - {v^2}}}dv{\text{ by the partial fractions method}}} \cr
& \,\,\,\,\,\frac{{3v - 1}}{{4{v^3} - {v^2}}} = \frac{{3v - 1}}{{{v^2}\left( {4v - 1} \right)}} = \frac{A}{v} + \frac{B}{{{v^2}}} + \frac{C}{{4v - 1}} \cr
& \,\,\,\,\,3v - 1 = Av\left( {4v - 1} \right) + B\left( {4v - 1} \right) + C{v^2}\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& \,\,\,\,\,{\text{let}}\,\,\,v = 0{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \,\,\,\,\,\,3\left( 0 \right) - 1 = A\left( 0 \right) + B\left( { - 1} \right) + C\left( 0 \right) \cr
& \,\,\,\,\,B = 1 \cr
& \,\,\,\,{\text{let}}\,\,\,v = 1/4{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \,\,\,\,\,\,3\left( {\frac{1}{4}} \right) - 1 = A\left( 0 \right) + B\left( 0 \right) + C\left( {\frac{1}{{16}}} \right) \cr
& \,\,\,\,\,C = - 4 \cr
& \,\,\,\,\,\,\,{\text{let}}\,\,\,v = 1,\,\,B = 1{\text{ and }}C = - 4{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \,\,\,\,\,3\left( 1 \right) - 1 = A\left( 1 \right)\left( 3 \right) + \left( 1 \right)\left( 3 \right) + \left( { - 4} \right){\left( 1 \right)^2}\,\,\, \cr
& \,\,\,\,\,A = 1 \cr
& \,\,\,\,{\text{Then}}{\text{,}}\,\,\,\frac{{3v - 1}}{{{v^2}\left( {4v - 1} \right)}} = \frac{1}{v} + \frac{1}{{{v^2}}} - \frac{4}{{4v - 1}} \cr
& \,\,\,\,\,\,\,\int {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} dv = \int {\frac{1}{v}} dv + \int {\frac{1}{{{v^2}}}} dv - \int {\frac{4}{{4v - 1}}} dv \cr
& \,\,\,\,\,\,\,\int {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} dv = \ln \left| v \right| - \frac{1}{v} - \ln \left| {4v - 1} \right| + C \cr
& \,\,\,\,\,\,\,\int {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} dv = \ln \left| {\frac{v}{{4v - 1}}} \right| - \frac{1}{v} + C \cr
& \cr
& \cr
& \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{3v - 1}}{{4{v^3} - {v^2}}}} = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {\frac{v}{{4v - 1}}} \right| - \frac{1}{v}} \right)_1^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {\frac{b}{{4b - 1}}} \right| - \frac{1}{b}} \right) - \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {\frac{1}{{4\left( 1 \right) - 1}}} \right| - \frac{1}{1}} \right) \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {\frac{1}{{4 - 1/b}}} \right| - \frac{1}{b}} \right) - \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {\frac{1}{3}} \right| - 1} \right) \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& = \ln \left| {\frac{1}{{4 - 0}}} \right| - 0 - \ln \left( {\frac{1}{3}} \right) + 1 \cr
& = \ln \left( {\frac{1}{4}} \right) - \ln \left( {\frac{1}{3}} \right) + 1 \cr
& = \ln \left( {\frac{1}{4}} \right) - \ln \left( {\frac{1}{3}} \right) + 1 \cr
& = \ln \left( {\frac{3}{4}} \right) + 1 \cr} $$
