How do you integrate $int(x+1)/((x-9)(x+8)(x-2))$ using partial fractions?

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Answer 1 · 2017-07-10T17:00:43

The answer is $=10/119ln(|x-9|)-7/170ln(|x+8|)-3/70ln(|x-2|)+C$

Let's perform the decomposition into partial fractions

$(x+1)/((x-9)(x+8)(x-2))=A/(x-9)+B/(x+8)+C/(x-2)$

$=(A(x-2)(x+8)+B(x-9)(x-2)+C(x-9)(x+8))/((x-9)(x+8)(x-2))$

The denominator is the same, we compare the numerator.

$x+1=A(x-2)(x+8)+B(x-9)(x-2)+C(x-9)(x+8)$

Let $x=9$ , $=>$ , $10=17*7A$ , $=>$ , $A=10/119$

Let $x=-8$ , $=>$ , $-7=-17*-10B$ , $=>$ , $B=-7/170$

Let $x=2$ , $=>$ , $3=-7*10C$ , $=>$ , $C=-3/70$

Therefore,

$(x+1)/((x-9)(x+8)(x-2))=(10/119)/(x-9)-(7/170)/(x+8)-(3/70)/(x-2)$

$int((x+1)dx)/((x-9)(x+8)(x-2))=int(10/119dx)/(x-9)-int(7/170dx)/(x+8)-int(3/70dx)/(x-2)$

$=10/119ln(|x-9|)-7/170ln(|x+8|)-3/70ln(|x-2|)+C$

How do you integrate int(x+1)/((x-9)(x+8)(x-2))int(x+1)/((x-9)(x+8)(x-2)) using partial fractions?