How do you integrate #int (1-x^2)/((x-9)(x-2)(x-2)) # using partial fractions?

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Answer 1 · 2017-11-14T19:45:25

#31/49*Ln(x-2)-80/49*ln(x-9)-3/7*(x-2)^(-1)+C#

I partitioned integrand into basic fractions,

#(1-x^2)/[(x-9)*(x-2)^2]=A/(x-9)+B/(x-2)+C/(x-2)^2#

After expanding denominator,

#A*(x-2)^2+B*(x-2)(x-9)+C*(x-9)=1-x^2#

After setting #x=2#, #-7C=-3#, so #C=3/7#

After setting #x=9#, #49A=-80#, so #A=-80/49#

After setting #x=1#, #A+8B-8C=0#, so #B=1/8*(8C-A)=31/49#

Thus,

#int (1-x^2)/[(x-9)*(x-2)^2]*dx#

=#-80/49*int (dx)/(x-9)+31/49*int (dx)/(x-2)+3/7*int (dx)/(x-2)^2#

=#31/49*Ln(x-2)-80/49*ln(x-9)-3/7*(x-2)^(-1)+C#