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