Question

How do you integrate `int(x+1)/((x+5)(x+3)(x+4))` using partial fractions?

Answer 1

The answer is `=-2ln|(x+5)|-ln|(x+3)|+3ln|(x+4)|+C`

Explanation:

Let's perform the decomposition into partial fractions

`(x+1)/((x+5)(x+3)(x+4))=A/(x+5)+B/(x+3)+C/(x+4)`

`=(A(x+3)(x+4)+B(x+5)(x+4)+C(x+5)(x+3))/((x+5)(x+3)(x+4))`

The denominators are the same, we compare the numerators

`(x+1)=(A(x+3)(x+4)+B(x+5)(x+4)+C(x+5)(x+3))`

Let `x=-5`, `=>`,

`-4=-2*-1*A`, `=>`, `A=-2`

Let `x=-3`, `=>`,

`-2=2*1*B`, `=>`, `B=-1`

Let `x=-4`, `=>`,

`-3=1*-1*C`, `=>`, `C=3`

So,

`(x+1)/((x+5)(x+3)(x+4))=-2/(x+5)-1/(x+3)+3/(x+4)`

`int((x+1)dx)/((x+5)(x+3)(x+4))=-2intdx/(x+5)-1intdx/(x+3)+3intdx/(x+4)`

`=-2ln|(x+5)|-ln|(x+3)|+3ln|(x+4)|+C`