Answer
$$\frac{1}{5}\ln \left| {4x - 1} \right| - \frac{1}{{10}}\ln \left( {4{x^2} + 1} \right) - \frac{1}{{10}}{\tan ^{ - 1}}2x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{16{x^3} - 4{x^2} + 4x - 1}}} \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{1}{{16{x^3} - 4{x^2} + 4x - 1}} = \frac{1}{{\left( {4x - 1} \right)\left( {4{x^2} + 1} \right)}} \cr
& \frac{1}{{\left( {4x - 1} \right)\left( {4{x^2} + 1} \right)}} = \frac{A}{{4x - 1}} + \frac{{Bx + C}}{{4{x^2} + 1}} \cr
& 1 = A\left( {4{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {4x - 1} \right) \cr
& 1 = 4A{x^2} + A + 4B{x^2} - Bx + 4Cx - C \cr
& {\text{Collect like terms}} \cr
& 1 = \left( {4A{x^2} + 4B{x^2}} \right) + \left( { - Bx + 4Cx} \right) + A - C \cr
& 4A + 4B = 0,{\text{ }} - B + 4C = 0,{\text{ }}A - C = 1 \cr
& {\text{Solving we obtain}} \cr
& A = \frac{4}{5},{\text{ }}B = - \frac{4}{5},{\text{ }}C = - \frac{1}{5} \cr
& \frac{1}{{\left( {4x - 1} \right)\left( {4{x^2} + 1} \right)}} = \frac{4}{{5\left( {4x - 1} \right)}} - \frac{{4x + 1}}{{5\left( {4{x^2} + 1} \right)}} \cr
& \int {\frac{{dx}}{{16{x^3} - 4{x^2} + 4x - 1}}} = \int {\frac{4}{{5\left( {4x - 1} \right)}}dx} - \int {\frac{{4x + 1}}{{5\left( {4{x^2} + 1} \right)}}} dx \cr
& = \frac{1}{5}\int {\frac{4}{{4x - 1}}dx} - \frac{1}{5}\int {\frac{{4x}}{{4{x^2} + 1}}} dx - \frac{1}{5}\int {\frac{1}{{4{x^2} + 1}}} dx \cr
& = \frac{1}{5}\int {\frac{4}{{4x - 1}}dx} - \frac{1}{{10}}\int {\frac{{8x}}{{4{x^2} + 1}}} dx - \frac{1}{{10}}\int {\frac{2}{{4{x^2} + 1}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{5}\ln \left| {4x - 1} \right| - \frac{1}{{10}}\ln \left( {4{x^2} + 1} \right) - \frac{1}{{10}}{\tan ^{ - 1}}2x + C \cr} $$
