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

Question

1478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-13T12:24:18

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

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

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