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

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Answer 1 · 2016-09-02T16:56:38

This could be integrated by substitution, but the question specifies partial fractions, so see below.

Factor the denominator:

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

solve for #A# and #B#

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

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

#Ax-5A+Bx+5B = 2x+0#

#A+B = 2#
#-5A+5B = 0#

#A = B = 1#

#int (2x)/(x^2-25) dx = int (1/(x+5)+1/(x-5)) dx#

# = ln abs(x+5)+ln abs(x-5)+C#

# = ln abs(x^2-25) +C#