Begin with the set up of the variables A, B, C, D
#(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+(Cx+D)/(x^2+1)#
the LCD(Least common Denominator) is #=(x^2+1)^2#
so that the right side of the equation becomes
#(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+((Cx+D)(x^2+1))/(x^2+1)^2#
expand
#(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+(Cx^3+Cx+Dx^2+D)/(x^2+1)^2#
combine
#(2x^2+5)/(x^2+1)^2=(Ax+B+Cx^3+Cx+Dx^2+D)/(x^2+1)^2#
rearrange
#(0*x^3+2x^2+0*x+5*x^0)/(x^2+1)^2=(Cx^3+Dx^2+(A+C)x+(B+D)x^0)/(x^2+1)^2#
the equations can now be found by equating the numerical coefficients:
#C=0#
#D=2#
#A+C=0#
#B+D=5#
Solving for A, B, C, D:
#A=0# and #B=3# and #C=0# and #D=2#
so that the partial fractions are
#(2x^2+5)/(x^2+1)^2=(Ax+B)/(x^2+1)^2+(Cx+D)/(x^2+1)=(0*x+3)/(x^2+1)^2+(0*x+2)/(x^2+1)#
and
#(2x^2+5)/(x^2+1)^2=3/(x^2+1)^2+2/(x^2+1)#
have a nice day! from the Philippines..
