How do you integrate #int (x-3x^2)/((x+9)(x-5)(x-7)) # using partial fractions?

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Answer 1 · 2017-10-24T11:33:51

#ln(((x+9)^(9/8)(x-5)^(5/2))/(x-7)^(7/4)) + C#

First you'll need to split up the integrand into partial fractions.

#(x-3x^2)/((x+9)(x-5)(x-7)) = A/(x+9)+B/(x-5)+C/(x-7)#

#x-3x^2 = A(x-5)(x-7) + B(x+9)(x-7) + C(x+9)(x-5)#

We'll substitute in the roots, #x=-9,5,7#, to find each of the numerator constants.

#x = -9#

#(-9)-3xx(-9)^2 = A xx (-9-5) xx (-9-7)#

#-252 = A xx -14 xx -16#

#A = 9/8#

#x = 5#

#(5)-3(5)^2 = B xx (5+9) xx (5-7)#

#-70 = B xx 14 xx -2#

#B = 5/2#

#x = 7#

#(7)-3(7)^2 = C xx (7 + 9) xx (7-2)#

#-140 = C xx 16 xx 5#

#C = -7/4#

Therefore, we now have that

#(x-3x^2)/((x+9)(x-5)(x-7)) = (9/8)/(x+9)+(5/2)/(x-5)-(7/4)/(x-7)#

Integrating #a/(x+b)# gives you #aln(x+b)#, so

#int (9/8)/(x+9)+(5/2)/(x-5)-(7/4)/(x-7)dx =#

#9/8ln(x+9) + 5/2ln(x-5) - 7/4ln(x-7) + C#

#= ln(((x+9)^(9/8)(x-5)^(5/2))/(x-7)^(7/4)) + C#