Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.5 Exercises - Page 549: 20

$$\frac{1}{{16}}\ln \left| {4{x^2} - 1} \right| - \frac{1}{{16}}\ln \left( {4{x^2} + 1} \right) + C$$

$$\eqalign{ & \int {\frac{x}{{16{x^4} - 1}}} dx \cr & {\text{Factor the denominator}} \cr & = \int {\frac{x}{{\left( {4{x^2} + 1} \right)\left( {4{x^2} - 1} \right)}}} dx \cr & = \int {\frac{x}{{\left( {2x + 1} \right)\left( {2x - 1} \right)\left( {4{x^2} + 1} \right)}}} dx \cr & {\text{Decompose the integrand into partial fractions}} \cr & \frac{x}{{\left( {2x + 1} \right)\left( {2x - 1} \right)\left( {4{x^2} + 1} \right)}} = \frac{A}{{2x + 1}} + \frac{B}{{2x - 1}} + \frac{{Cx + D}}{{4{x^2} + 1}} \cr & x = A(2x - 1)(4{x^2} + 1) + B(2x + 1)(4{x^2} + 1) + \left( {Cx + D} \right)\left( {{x^2} - 4} \right) \cr & {\text{Expanding}} \cr & x = A(8{x^3} + 2x - 4{x^2} - 1) + B(8{x^3} + 2x + 4{x^2} + 1) \cr & + 4C{x^3} - Cx + 4D{x^2} - D \cr & {\text{Group like terms}} \cr & x = 8A{x^3} + 2Ax - 4A{x^2} - A + 8B{x^3} + 2Bx + 4B{x^2} + B + 4C{x^3} \cr & - Cx + 4D{x^2} - D \cr & x = (8A{x^3} + 8B{x^3} + 4C{x^3}) + ( - 4A{x^2} + 4B{x^2} + 4D{x^2}) \cr & + \left( {2Ax + 2Bx - Cx} \right) + \left( { - A + B - D} \right) \cr & {\text{Setting up a system of equations}} \cr & 8A + 8B + 4C = 0 \cr & - 4A + 4B + 4D = 0 \cr & 2A + 2A - C = 1 \cr & - A + B - D = 0 \cr & {\text{Solving the system of equations by using a CAS or calculator}} \cr & A = \frac{1}{8},{\text{ }}B = \frac{1}{8},{\text{ }}C = - \frac{1}{2},{\text{ }}D = 0 \cr & \frac{x}{{\left( {2x + 1} \right)\left( {2x - 1} \right)\left( {4{x^2} + 1} \right)}} = \frac{{1/8}}{{2x + 1}} + \frac{{1/8}}{{2x - 1}} - \frac{{1/2x}}{{4{x^2} + 1}} \cr & \int {\left( {\frac{{1/8}}{{2x + 1}} + \frac{{1/8}}{{2x - 1}} - \frac{{1/2}}{{4{x^2} + 1}}} \right)} dx \cr & {\text{Integrating}} \cr & = \frac{1}{{16}}\ln \left| {2x + 1} \right| + \frac{1}{{16}}\left| {2x - 1} \right| - \frac{1}{{16}}\ln \left( {4{x^2} + 1} \right) + C \cr & {\text{Using logarithmic properties}} \cr & = \frac{1}{{16}}\ln \left| {\left( {2x + 1} \right)\left( {2x - 1} \right)} \right| - \frac{1}{{16}}\ln \left( {4{x^2} + 1} \right) + C \cr & = \frac{1}{{16}}\ln \left| {4{x^2} - 1} \right| - \frac{1}{{16}}\ln \left( {4{x^2} + 1} \right) + C \cr} $$