Chapter 7 - Principles Of Integral Evaluation - 7.5 Integrating Rational Functions By Partial Fractions - Exercises Set 7.5 - Page 522: 29

$$ - \frac{7}{{34}}\ln \left| {4x - 1} \right| + \frac{6}{{17}}\ln \left( {{x^2} + 1} \right) + \frac{3}{{17}}\arctan \left( x \right) + C$$

$$\eqalign{ & \int {\frac{{2{x^2} - 1}}{{\left( {4x - 1} \right)\left( {{x^2} + 1} \right)}}} dx \cr & {\text{The partial fraction decomposition of the integrand is}} \cr & \frac{{2{x^2} - 1}}{{\left( {4x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{4x - 1}} + \frac{{Bx + C}}{{{x^2} + 1}} \cr & {\text{Multiplying by }}\left( {4x - 1} \right)\left( {{x^2} + 1} \right){\text{ yields}} \cr & 2{x^2} - 1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {4x - 1} \right) \cr & 2{x^2} - 1 = A{x^2} + A + 4B{x^2} - Bx + 4Cx - C \cr & {\text{Collecting like powers of }}x{\text{, this becomes}} \cr & 2{x^2} - 1 = \left( {A{x^2} + 4B{x^2}} \right) + \left( { - Bx + 4Cx} \right) + \left( {A - C} \right) \cr & {\text{Equating corresponding coefficients yields the following system }} \cr & {\text{of linear equations}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,A + 4B = 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, - B + 4C = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,A - C = - 1 \cr & {\text{Solving the system of linear equations by a calculator we obtain}} \cr & \,\,\,A = - \frac{{14}}{{17}},\,\,\,B = \frac{{12}}{{17}},\,\,\,C = \frac{3}{{17}} \cr & \cr & {\text{Then}}{\text{, the integrand can be written as}} \cr & \frac{{2{x^2} - 1}}{{\left( {4x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{ - 14/17}}{{4x - 1}} + \frac{{\left( {12/17} \right)x + 3/17}}{{{x^2} + 1}} \cr & \int {\frac{{2{x^2} - 1}}{{\left( {4x - 1} \right)\left( {{x^2} + 1} \right)}}} dx = \int {\left( {\frac{{ - 14/17}}{{4x - 1}} + \frac{{\left( {12/17} \right)x + 3/17}}{{{x^2} + 1}}} \right)} dx \cr & = \int {\left( { - \frac{{14}}{{17\left( {4x - 1} \right)}} + \frac{{12}}{{17}}\left( {\frac{x}{{{x^2} + 1}}} \right) + \frac{3}{{17}}\left( {\frac{1}{{{x^2} + 1}}} \right)} \right)} dx \cr & {\text{Integrating}} \cr & = - \frac{{14}}{{17}}\left( {\frac{1}{4}\ln \left| {4x - 1} \right|} \right) + \frac{{12}}{{17}}\left( {\frac{1}{2}\ln \left( {{x^2} + 1} \right)} \right) + \frac{3}{{17}}\arctan \left( x \right) + C \cr & = - \frac{7}{{34}}\ln \left| {4x - 1} \right| + \frac{6}{{17}}\ln \left( {{x^2} + 1} \right) + \frac{3}{{17}}\arctan \left( x \right) + C \cr} $$