## Thomas' Calculus 13th Edition

$$x + \ln \left| {\frac{{x - 1}}{x}} \right| + C$$
\eqalign{ & \int {\frac{{{x^3} + 1}}{{{x^3} - x}}} dx \cr & {\text{Perform the long division}} \cr & \frac{{{x^3} + 1}}{{{x^3} - x}} = 1 + \frac{{x + 1}}{{{x^3} - x}} \cr & \cr & {\text{Then}}{\text{,}} \cr & \int {\frac{{{x^3} + 1}}{{{x^3} - x}}} dx = \int {\left( {1 + \frac{{x + 1}}{{{x^3} - x}}} \right)} dx \cr & = \int {dx} + \int {\frac{{x + 1}}{{{x^3} - x}}} dx \cr & = \int {dx} + \int {\frac{{x + 1}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}}} dx \cr & = \int {dx} + \int {\frac{1}{{x\left( {x - 1} \right)}}} dx \cr & {\text{Decompose }}\frac{1}{{x\left( {x - 1} \right)}}{\text{ into partial fractions}} \cr & \frac{1}{{x\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} \cr & {\text{Multiply by }}x\left( {x - 1} \right){\text{ and simplify}} \cr & 1 = A\left( {x - 1} \right) + Bx \cr & {\text{if }}x = 0,\,\,\,A = - 1 \cr & {\text{if }}x = 1,\,\,\,B = 1 \cr & \cr & {\text{Replace the coefficients}} \cr & \frac{1}{{x\left( {x - 1} \right)}} = - \frac{1}{x} + \frac{1}{{x - 1}} \cr & \cr & {\text{Then}}{\text{,}} \cr & \int {dx} + \int {\frac{1}{{x\left( {x - 1} \right)}}} dx = \int {dx} + \int {\left( { - \frac{1}{x} + \frac{1}{{x - 1}}} \right)} dx \cr & {\text{Integrate}} \cr & = x - \ln \left| x \right| + \ln \left| {x - 1} \right| + C \cr & = x + \ln \left| {\frac{{x - 1}}{x}} \right| + C \cr}