How do you integrate int (1-x^2)/((x-9)(x-2)(x-2)) ∫1−x2(x−9)(x−2)(x−2) using partial fractions?

1 Answer
Cem Sentin
Nov 14, 2017

31/49*Ln(x-2)-80/49*ln(x-9)-3/7*(x-2)^(-1)+C3149⋅ln(x−2)−8049⋅ln(x−9)−37⋅(x−2)−1+C

Explanation:

I partitioned integrand into basic fractions,

(1-x^2)/[(x-9)*(x-2)^2]=A/(x-9)+B/(x-2)+C/(x-2)^21−x2(x−9)⋅(x−2)2=Ax−9+Bx−2+C(x−2)2

After expanding denominator,

A*(x-2)^2+B*(x-2)(x-9)+C*(x-9)=1-x^2A⋅(x−2)2+B⋅(x−2)(x−9)+C⋅(x−9)=1−x2

After setting x=2x=2, -7C=-3−7C=−3, so C=3/7C=37

After setting x=9x=9, 49A=-8049A=−80, so A=-80/49A=−8049

After setting x=1x=1, A+8B-8C=0A+8B−8C=0, so B=1/8*(8C-A)=31/49B=18⋅(8C−A)=3149

Thus,

int (1-x^2)/[(x-9)*(x-2)^2]*dx∫1−x2(x−9)⋅(x−2)2⋅dx

=-80/49*int (dx)/(x-9)+31/49*int (dx)/(x-2)+3/7*int (dx)/(x-2)^2−8049⋅∫dxx−9+3149⋅∫dxx−2+37⋅∫dx(x−2)2

=31/49*Ln(x-2)-80/49*ln(x-9)-3/7*(x-2)^(-1)+C3149⋅ln(x−2)−8049⋅ln(x−9)−37⋅(x−2)−1+C

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