How do you integrate $int(x+1)/((x+5)(x-1)(x-4))$ using partial fractions?

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Answer 1 · 2017-09-01T19:00:02

The integral is $=-2/27ln(|x+5|)-1/9ln(|x-1|)+5/27ln(|x-4|)+C$

Let's perform the decomposition into partial fractions

$(x+1)/((x+5)(x-1)(x-4))=A/(x+5)+B/(x-1)+C/(x-4)$

$=(A(x-1)(x-4)+B(x+5)(x-4)+C(x+5)(x-1))/((x+5)(x-1)(x-4))$

The denominators are the same, we compare the numerators

$x+1=A(x-1)(x-4)+B(x+5)(x-4)+C(x+5)(x-1)$

Let $x=-5$ , $=>$ , $-4=A*(-6)(-9)$ , $=>$ , $A=-2/27$

Let $x=1$ , $=>$ , $2=B*(6)(-3)$ , $=>$ , $B=-1/9$

Let $x=4$ , $=>$ , $5=C*(9)(3)$ , $=>$ , $C=5/27$

Therefore,

$(x+1)/((x+5)(x-1)(x-4))=(-2/27)/(x+5)+(-1/9)/(x-1)+(5/27)/(x-4)$

$int((x+1)dx)/((x+5)(x-1)(x-4))=int(-2/27dx)/(x+5)+int(-1/9dx)/(x-1)+int(5/27dx)/(x-4)$

$=-2/27ln(|x+5|)-1/9ln(|x-1|)+5/27ln(|x-4|)+C$

How do you integrate int(x+1)/((x+5)(x-1)(x-4))int(x+1)/((x+5)(x-1)(x-4)) using partial fractions?