Answer
$$\frac{{{x^3}}}{3} + x + \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^4}}}{{{x^2} - 1}}} dx \cr
& \cr
& {\text{Apply long division to the integrand}} \cr
& \frac{{{x^4}}}{{{x^2} - 1}} = {x^2} + 1 + \frac{1}{{{x^2} - 1}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{{x^4}}}{{{x^2} - 1}}} dx = \int {\left( {{x^2} + 1 + \frac{1}{{{x^2} - 1}}} \right)} dx \cr
& = \int {\left( {{x^2} + 1} \right)} dx + \int {\frac{1}{{{x^2} - 1}}} dx \cr
& = \frac{{{x^3}}}{3} + x + \int {\frac{1}{{{x^2} - 1}}} dx \cr
& \cr
& {\text{Solve the integral by the method of partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{{x^2} - 1}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} \cr
& {\text{Multiplying by }}\left( {x - 1} \right)\left( {x + 1} \right){\text{, we have}} \cr
& {\text{1}} = A\left( {x + 1} \right) + B\left( {x - 1} \right) \cr
& {\text{if we set }}x = 1, \cr
& {\text{1}} = A\left( 2 \right) \cr
& A = 1/2 \cr
& {\text{if we set }}x = - 1 \cr
& {\text{1}} = A\left( 0 \right) + B\left( { - 2} \right) \cr
& B = - 1/2 \cr
& {\text{then}} \cr
& \frac{1}{{{x^2} - 1}} = \frac{A}{x} + \frac{B}{{x - 1}} = \frac{{1/2}}{{x - 1}} - \frac{{1/2}}{{x + 1}} \cr
& \int {\frac{1}{{{x^2} - 1}}} dx = \int {\left( {\frac{{1/2}}{{x - 1}} - \frac{{1/2}}{{x + 1}}} \right)} dx \cr
& \int {\frac{1}{{{x^2} - 1}}} dx = \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + C \cr
& \int {\frac{1}{{{x^2} - 1}}} dx = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C \cr
& \cr
& \frac{{{x^3}}}{3} + x + \int {\frac{1}{{{x^2} - 1}}} dx = \frac{{{x^3}}}{3} + x + \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C \cr} $$
