How do you integrate #6/(x^2-25)^2# using partial fractions?

1 Answer
Jun 26, 2018

# 3/250[ln|(x+5)/(x-5)|-10/(x^2-25)]+C#

Explanation:

#6/(x^2-25)^2 = 6/((x-5)^2(x+5)^2)#

You can carry out the partial fractions decomposition by following the standard method, i.e., by starting with the form

#6/((x-5)^2(x+5)^2) = A/(x-5)+B/(x+5)#
#qquadqquadqquadqquadqquadqquad qquadqquad+C/(x-5)^2+D/(x+5)^2 #

and rewriting this as

#6 = A(x-5)(x+5)^2+B(x-5)^2(x+5)#

#qquadqquad +C(x+5)^2+D(x-5)^2#

  • substituting #x=+5# gives #100C=6implies C = 3/50#
  • substituting #x=-5# gives #100D=6implies D = 3/50#
  • comparing coefficients of #x^3# on both sides yield #A +B = 0#
  • comparing coefficients of #x^0# on both sides yield
    #-5^3A +5^3B +5^2C+(-5)^2D= 6 implies#
    #A-B = (C+D)/5-6/5^3 = -3/125implies#
    #A = -B = -3/250#

Thus

#6/(x^2-25)^2 = -3/(250(x-5))+3/(250(x+5))#
#qquadqquadqquadqquadqquadqquad qquadqquad+3/(50(x-5)^2)+3/(50(x+5)^2) #

Thus the required integral is

#int (6dx)/(x^2-25)^2 = -int (3dx)/(250(x-5))+int(3dx)/(250(x+5))#
#qquadqquadqquadqquadqquadqquad qquadqquad+int (3dx)/(50(x-5)^2)+int (3dx)/(50(x+5)^2) #
#qquadqquad = 3/250 ln|(x+5)/(x-5)|-3/50[1/(x-5)+1/(x+5)]+C#
#qquadqquad = 3/250[ln|(x+5)/(x-5)|-10/(x^2-25)]+C#