How do you integrate $int(x)/((x+4)(x+6)(x+1))$ using partial fractions?

Question

1269 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-15T08:45:42

$2/3 ln (x+4) -3/5 ln (x+6) -1/15 ln (x+1)+C$

C=the constant of inegration

$x=A(x^2+7x+6)+B(x^2+5x+4)+C(x^2+10x+24)$

$A=2/3$
$B=-3/5$
$C=-1/15$
$int x/((x+4)(x+6)(x+1)) dx = int A/(x+4) dx+ int B/(x+6) dx+int C/(x+1) dx$

How do you integrate int(x)/((x+4)(x+6)(x+1))int(x)/((x+4)(x+6)(x+1)) using partial fractions?