Answer
$$\ln \left| x \right| - \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{3}\ln | {{x^2} - x + 1}| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^4} + x}}} dx \cr
& {\text{Decompose the integrand }}\frac{1}{{{x^4} + x}}{\text{ into partial fractions}} \cr
& \frac{1}{{x\left( {{x^3} + 1} \right)}} = \frac{1}{{x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 1}} + \frac{{Cx + D}}{{{x^2} - x + 1}} \cr
& {\text{multiply by }}x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right){\text{ and simplify}} \cr
& 1 = A\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) + Bx\left( {{x^2} - x + 1} \right) + \left( {Cx + D} \right)x\left( {x + 1} \right) \cr
& {\text{Expanding the equation }} \cr
& 1 = A\left( {{x^3} + 1} \right) + B{x^3} - B{x^2} + Bx + C{x^3} + C{x^2} + D{x^2} + Dx \cr
& \cr
& {\text{Group terms}} \cr
& 1 = \left( {A{x^3} + B{x^3} + C{x^3}} \right) + \left( { - B{x^2} + C{x^2} + D{x^2}} \right) + Bx + Dx + A \cr
& {\text{Equating coefficients}} \cr
& \,\,\,\,\,\,\,\,A + B + C = 0 \cr
& \,\,\,\,\,\,\,\,\, - B + C + D = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,B + D = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A = 1 \cr
& {\text{Solving the system of equations by using a calculator}}{\text{, we obtain}} \cr
& A = 1,\,\,\,B = - \frac{1}{3},\,\,\,C = - \frac{2}{3},\,\,\,D = \frac{1}{3} \cr
& \cr
& \frac{1}{{x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{1}{x} + \frac{{ - 1/3}}{{x + 1}} + \frac{{\left( { - 2/3} \right)x + 1/3}}{{{x^2} - x + 1}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{1}{{{x^4} + x}}} dx = \int {\left( {\frac{1}{x} - \frac{1}{{3\left( {x + 1} \right)}} - \frac{1}{3}\left( {\frac{{2x - 1}}{{{x^2} - x + 1}}} \right)} \right)} dx \cr
& \cr
& {\text{Integrating, we get }} \cr
& = \ln \left| x \right| - \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{3}\ln | {{x^2} - x + 1} | + C \cr} $$
