How do you integrate $int(x+1)/((x+5)(x+2)(x-5))$ using partial fractions?

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Answer 1 · 2017-01-10T14:28:28

The answer is $=-2/15(∣x+5∣)+1/21ln(∣x+2∣)+3/35ln(∣x-5∣)+C$

Perform the decomposition into partial fractions

$(x+1)/((x+5)(x+2)(x-5))=A/(x+5)+B/(x+2)+C/(x-5)$

$=(A(x+2)(x-5)+B(x+5)(x-5)+C(x+5)(x+2))/((x+5)(x+2)(x-5))$

Therefore, by equating the numerators

$(x+1)=(A(x+2)(x-5)+B(x+5)(x-5)+C(x+5)(x+2))$

Let $x=-5$ , $=>$ , $-4=30A$ , $=>$ , $A=-2/15$

Let $x=-2$ , $=>$ , $-1=-21B$ , $=>$ , $B=1/21$

Let $x=5$ , $=>$ , $6=70C$ , $=>$ , $C=3/35$

so,

$(x+1)/((x+5)(x+2)(x-5))=(-2/15)/(x+5)+(1/21)/(x+2)+(3/35)/(x-5)$

Now, perform the integration

$int((x+1)dx)/((x+5)(x+2)(x-5))=-2/15intdx/(x+5)+1/21intdx/(x+2)+3/35intdx/(x-5)$

$=-2/15(∣x+5∣)+1/21ln(∣x+2∣)+3/35ln(∣x-5∣)+C$

How do you integrate int(x+1)/((x+5)(x+2)(x-5))int(x+1)/((x+5)(x+2)(x-5)) using partial fractions?