Answer
$$\frac{1}{4}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{1}{2}{\tan ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{{x^4} - 1}}} dx \cr
& {\text{Decompose the integrand }}\frac{{{x^2}}}{{{x^4} - 1}}{\text{ into partial fractions}} \cr
& \frac{{{x^2}}}{{{x^4} - 1}} = \frac{{{x^2}}}{{\left( {{x^2} + 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}} \cr
& \frac{{{x^2}}}{{{x^4} - 1}} = \frac{A}{{x + 1}} + \frac{B}{{x - 1}} + \frac{{Cx + D}}{{{x^2} + 1}} \cr
& {\text{Multiply by }}\left( {{x^2} + 1} \right)\left( {x + 1} \right)\left( {x - 1} \right){\text{ and simplify}} \cr
& {x^2} = A\left( {x - 1} \right)\left( {{x^2} + 1} \right) + B\left( {x + 1} \right)\left( {{x^2} + 1} \right) + \left( {Cx + D} \right)\left( {{x^2} - 1} \right) \cr
& {x^2} = A{x^3} - A{x^2} + Ax - A + B{x^3} + B{x^2} + Bx + B + C{x^3} - Cx + D{x^2} - D \cr
& {\text{Group terms}} \cr
& {x^2} = \left( {A{x^3} + B{x^3} + C{x^3}} \right) + \left( { - A{x^2} + B{x^2} + D{x^2}} \right) + \left( {Ax + Bx - Cx} \right) + \left( { - A + B - D} \right) \cr
& {\text{Equate coefficients}} \cr
& A + B + C = 0 \cr
& - A + B + D = 1 \cr
& A + B - C = 0 \cr
& - A + B - D = 0 \cr
& {\text{Solve the system of equations}} \cr
& A = - 1/4,\,\,\,\,B = 1/4,\,\,\,C = 0,\,\,\,D = 1/2 \cr
& {\text{Replace the coefficients}} \cr
& \frac{{{x^2}}}{{{x^4} - 1}} = \frac{{ - 1/4}}{{x + 1}} + \frac{{1/4}}{{x - 1}} + \frac{{1/2}}{{{x^2} + 1}} \cr
& \int {\frac{{{x^2}}}{{{x^4} - 1}}} dx = \int {\left( {\frac{{ - 1/4}}{{x + 1}} + \frac{{1/4}}{{x - 1}} + \frac{{1/2}}{{{x^2} + 1}}} \right)} dx \cr
& {\text{Integrate}} \cr
& = - \frac{1}{4}\ln \left| {x + 1} \right| + \frac{1}{4}\ln \left| {x - 1} \right| + \frac{1}{2}{\tan ^{ - 1}}x + C \cr
& = \frac{1}{4}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{1}{2}{\tan ^{ - 1}}x + C \cr} $$
