Answer
$$\ln \left( {\frac{{{x^2}}}{{{x^2} + 1}}} \right) + \frac{1}{{{x^2} + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{2}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{Decompose }}\frac{2}{{x{{\left( {{x^2} + 1} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{2}{{x{{\left( {{x^2} + 1} \right)}^2}}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& {\text{Multiply both sides by }}x{\left( {{x^2} + 1} \right)^2} \cr
& 2 = A{\left( {{x^2} + 1} \right)^2} + x\left( {{x^2} + 1} \right)\left( {Bx + C} \right) + x\left( {Dx + E} \right) \cr
& 2 = A\left( {{x^4} + 2{x^2} + 1} \right) + \left( {{x^3} + x} \right)\left( {Bx + C} \right) + D{x^2} + Ex \cr
& 2 = A{x^4} + 2A{x^2} + A + B{x^4} + C{x^3} + B{x^2} + Cx + D{x^2} + Ex \cr
& {\text{Group like terms}} \cr
& 2 = \left( {A + B} \right){x^4} + C{x^3} + \left( {2A + B + D} \right){x^2} + \left( {C + E} \right)x + A \cr
& {\text{We obtain the system of linear equations}} \cr
& A + B = 0,\,\,\,\,C = 0,\,\,\,\,2A + B + D = 0,\,\,\,\,C + E = 0,\,\,A = 2 \cr
& {\text{Solving the system we obtain}} \cr
& A = 2,\,\,\,B = - 2,\,\,\,\,C = 0,\,\,\,\,D = - 2,\,\,\,\,E = 0 \cr
& {\text{Therefore,}} \cr
& \frac{2}{{x{{\left( {{x^2} + 1} \right)}^2}}} = \frac{2}{x} - \frac{{2x}}{{{x^2} + 1}} - \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \cr
& \int {\frac{2}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx = \int {\left[ {\frac{2}{x} - \frac{{2x}}{{{x^2} + 1}} - \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]} dx \cr
& = \int {\frac{2}{x}} dx - \int {\frac{{2x}}{{{x^2} + 1}}} dx - \int {{{\left( {{x^2} + 1} \right)}^{ - 2}}\left( {2x} \right)dx} \cr
& {\text{Integrate}} \cr
& = 2\ln \left| x \right| - \ln \left( {{x^2} + 1} \right) - \frac{{{{\left( {{x^2} + 1} \right)}^{ - 1}}}}{{ - 1}} + C \cr
& = \ln {x^2} - \ln \left( {{x^2} + 1} \right) + \frac{1}{{{x^2} + 1}} + C \cr
& = \ln \left( {\frac{{{x^2}}}{{{x^2} + 1}}} \right) + \frac{1}{{{x^2} + 1}} + C \cr} $$
