Question

How do you integrate `int (x-9)/((x+8)(x-7)(x-5))` using partial fractions?

Answer 1

`-17/195 ln(x+8) -1/15 ln(x-7)+2/13 ln (x-5)+ C_1`

Explanation:

To Integrate, partial fractions are required, which can be done as follows: Let,
`(x-9)/((x+8)(x-7)(x-5))= A/(x+8) +B/(x-7) +C/(x-5)`

=`(A(x-7)(x-5) +B(x+8)(x-5) +C(x+8)(x-7))/((x+8)(x-7)(x-5))`

Thus`x-9= A(x^2-12x+35)+B(x^2 +3x-40)+C(x^2 +x-56)`

Comparing like coefficients, it would be
A+B+C=0, -12A +3B +C=1 and 35A -40B-56C=-9. Solving for A,B and C, it would be

`A= -17/195, B= -1/15 and C= 2/13`

The given integral would thus be `-17/195 ln(x+8) -1/15 ln(x-7)+2/13 ln (x-5)+ C_1`