How do you integrate `int (x+5)/((x+3)(x-7)(x-5))` using partial fractions?

from the given `int (x+5)/((x+3)(x-7)(x-5))` `dx` =

`intA/(x+3)``dx` + `B/(x-7)``dx` + `C/(x-5)`*`dx`

our equation becomes

`(x+5)/((x+3)(x-7)(x-5)`= `A/(x+3)+B/(x-7)+C/(x-5)`

it follows;

`(x+5)/((x+3)(x-7)(x-5)`=

`(A(x-7)(x-5)+B(x+3)(x-5)+C(x+3)(x-5))/((x+3)(x-5)(x-7))`

by using only the numerators;

`x+5=A(x^2-12x+35)+B(x^2-2x-15)+C(x^2-4x-21)`

collecting the like terms

`x+5=(A+B+C)x^2+(-12A-2B-4C)x+(35A-15B-21C)`

because the left side of the equation means

`0*x^2+1*x+5`

and the right side means

`(A+B+C)x^2+(-12A-2B-4C)x+(35A-15B-21C)`

the equations are now formed;

`A+B+C =0` first equation

`-12A-2B-4C=1` second equation

`35A-15B-21C=5` third equation

using your skills in solving 3 equations with 3 unknowns A, B, C

the values are `A=1/40`, `B=3/5`, and `C=-5/8`

Now, go back to the first line of the explanation to do the integration procedures.

`intA/(x+3)``dx` + `B/(x-7)``dx` + `C/(x-5)`*`dx`

`=int(1/40)/(x+3)``dx` + `int (3/5)/(x-7)``dx` + `int (-5/8)/(x-5)``dx`

final answer becomes

`int (x+5)/((x+3)(x-7)(x-5)` `dx`

`= 1/40*ln(x+3)+3/5*ln(x-7)-5/8*ln(x-5) + K`

for a definite integral, a constant K must be added.