I decomposed integrand into basic fractions,
#(x^2+1)/[x*(x+1)*(x^2+2)]#
=#A/x+B/(x+1)+(Cx+D)/(x^2+2)#
After expanding denominator,
#A(x+1)(x^2+2)+Bx(x^2+2)+(Cx+D)x(x+1)=x^2+1#
#A*(x^3+x^2+2x+2)+B*(x^3+2x)+(Cx+D)*(x^2+x)=x^2+1#
#(A+B+C)*x^3+(A+C+D)*x^2+(2A+2B+D)*x+2A=x^2+1#
After equating coefficients,
#A+B+C=0#, #A+C+D=1#, #2A+2B+D=0# and #2A=1#
From these equations, #A=1/2, B=-2/3, C=1/6 and D=1/3#
Thus,
#int (x^2+1)/[x*(x+1)*(x^2+2)]*dx#
=#1/2*int (dx)/x-2/3*int (dx)/(x+1)+1/6*int ((x+2)*dx)/(x^2+2)#
=#1/2*Lnx-2/3*Ln(x+1)+1/12*int ((2x+4)*dx)/(x^2+2)#
=#1/2*Lnx-2/3*Ln(x+1)+1/12*int (2x*dx)/(x^2+2)+1/3*int (dx)/(x^2+2)#
=#1/2*Lnx-2/3*Ln(x+1)+1/12*Ln(x^2+2)+sqrt2/6*int (sqrt2*dx)/(x^2+2)#
=#1/2*Lnx-2/3*Ln(x+1)+1/12*Ln(x^2+2)+sqrt2/6*arctan(x/sqrt2)+C#
