How do you integrate $int (1-2x^2)/((x-2)(x+6)(x+4))$ using partial fractions?

Question

1322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-12T10:20:38

$-7/48ln (x-2)-71/16ln(x+6)+31/12ln(x+4) + C$

from the given
$int (1-2x^2)/((x-2)(x+6)(x+4))$ $dx$

$A/(x-2)+B/(x+6)+C/(x+4)$ = $(1-2x^2)/((x-2)(x+6)(x+4))$

then

$A(x+6)(x+4)+B(x-2)(x+4)+C(x-2)(x+6)$

= $1 - 2x^2$

then

$A(x^2+10x+24)+B(x^2+2x-8)+C(x^2+4x-12)$

= $1-2x^2$

The 3 equations are now defined

$A+B+C=-2$

$10A+2B+4C=0$

$24A-8B-12C=1$

Use Algebra to find A, B, C

$A=-7/48$

$B=-71/16$

$C=31/12$

The integral becomes:

$int (-7/48)/(x-2)$ $dx$ + $int (-71/16)/(x+6)$ $dx$ + $int (31/12)/(x+4)$ $dx$

$-7/48ln(x-2)$ - $71/16ln(x+6)$ + $31/12ln(x+4)$

How do you integrate int (1-2x^2)/((x-2)(x+6)(x+4)) int (1-2x^2)/((x-2)(x+6)(x+4)) using partial fractions?