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

1 Answer
Narad T.
Sep 1, 2017

The integral is =-2/27ln(|x+5|)-1/9ln(|x-1|)+5/27ln(|x-4|)+C=−227ln(|x+5|)−19ln(|x−1|)+527ln(|x−4|)+C

Explanation:

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)x+1(x+5)(x−1)(x−4)=Ax+5+Bx−1+Cx−4

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

Let x=-5x=−5, =>⇒, -4=A*(-6)(-9)−4=A⋅(−6)(−9), =>⇒, A=-2/27A=−227

Let x=1x=1, =>⇒, 2=B*(6)(-3)2=B⋅(6)(−3), =>⇒, B=-1/9B=−19

Let x=4x=4, =>⇒, 5=C*(9)(3)5=C⋅(9)(3), =>⇒, C=5/27C=527

Therefore,

(x+1)/((x+5)(x-1)(x-4))=(-2/27)/(x+5)+(-1/9)/(x-1)+(5/27)/(x-4)x+1(x+5)(x−1)(x−4)=−227x+5+−19x−1+527x−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)∫(x+1)dx(x+5)(x−1)(x−4)=∫−227dxx+5+∫−19dxx−1+∫527dxx−4

=-2/27ln(|x+5|)-1/9ln(|x-1|)+5/27ln(|x-4|)+C=−227ln(|x+5|)−19ln(|x−1|)+527ln(|x−4|)+C

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