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,

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

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

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

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