Chapter 7 - Principles Of Integral Evaluation - 7.5 Integrating Rational Functions By Partial Fractions - Exercises Set 7.5 - Page 521: 22

$$\frac{{{x^3}}}{3} - \frac{1}{4}\ln \left| x \right| + \frac{1}{8}\ln \left| {{x^2} - 4} \right| + C$$

$$\eqalign{ & \int {\frac{{{x^5} - 4{x^3} + 1}}{{{x^3} - 4x}}} dx \cr & {\text{The integrand is an improper rational function since the numerator }} \cr & {\text{has degree 5 and the denominator 3}}{\text{. Thus}}{\text{, using the long division}} \cr & \,\,\,\,\,\,\frac{{{x^5} - 4{x^3} + 1}}{{{x^3} - 4x}} = {x^2} + \frac{1}{{{x^3} - 4x}} \cr & {\text{The integrand can be expressed as}} \cr & \int {\left( {{x^2} + \frac{1}{{{x^3} - 4x}}} \right)} dx \cr & = \frac{{{x^3}}}{3} + \int {\frac{1}{{{x^3} - 4x}}} dx \cr & {\text{Decomposing the integrand into partial fractions}} \cr & {\text{Factor the denominator}} \cr & \frac{1}{{{x^3} - 4x}} = \frac{1}{{x\left( {{x^2} - 4} \right)}} = \frac{1}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} \cr & \frac{1}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} \cr & {\text{Multiplying the equation by }}x\left( {x + 2} \right)\left( {x - 2} \right){\text{, we have}} \cr & 1 = A\left( {x + 2} \right)\left( {x - 2} \right) + Bx\left( {x - 2} \right) + Cx\left( {x + 2} \right) \cr & {\text{if we set }}x = 0 \cr & 1 = A\left( 2 \right)\left( { - 2} \right) + B\left( 0 \right) + C\left( 0 \right) \cr & A = - 1/4 \cr & {\text{if we set }}x = - 2 \cr & 1 = A\left( 0 \right) + B\left( { - 2} \right)\left( { - 2 - 2} \right) + C\left( 0 \right) \cr & B = 1/8 \cr & {\text{if we set }}x = 2 \cr & 1 = A\left( 0 \right) + B\left( 0 \right) + C\left( 2 \right)\left( {2 + 2} \right) \cr & C = 1/8 \cr & \cr & {\text{then}} \cr & \frac{1}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{{ - 1/4}}{x} + \frac{{1/8}}{{x + 2}} + \frac{{1/8}}{{x - 2}} \cr & \cr & \frac{{{x^3}}}{3} + \int {\frac{1}{{{x^3} - 4x}}} dx = \frac{{{x^3}}}{3} + \int {\left( {\frac{{ - 1/4}}{x} + \frac{{1/8}}{{x + 2}} + \frac{{1/8}}{{x - 2}}} \right)} dx \cr & {\text{Integrating}} \cr & = \frac{{{x^3}}}{3} - \frac{1}{4}\ln \left| x \right| + \frac{1}{8}\ln \left| {x + 2} \right| + \frac{1}{8}\ln \left| {x - 2} \right| + C \cr & = \frac{{{x^3}}}{3} - \frac{1}{4}\ln \left| x \right| + \frac{1}{8}\ln \left| {\left( {x + 2} \right)\left( {x - 2} \right)} \right| + C \cr & = \frac{{{x^3}}}{3} - \frac{1}{4}\ln \left| x \right| + \frac{1}{8}\ln \left| {{x^2} - 4} \right| + C \cr} $$