How do you integrate #6/(x^2-25)^2# using partial fractions?
1 Answer
Explanation:
You can carry out the partial fractions decomposition by following the standard method, i.e., by starting with the form
and rewriting this as
- 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
Thus the required integral is