How do you integrate $f(x)=(x^2-2x)/((x^2-3)(x-3)(x-8))$ using partial fractions?

Question

1471 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-05T16:42:09

$int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]$

= $5/122Ln(x-8)$ + $(21+5sqrt3)/732ln(x+sqrt3)$ - $1/10Ln(x-3)$ - $(21-5sqrt3)/6ln(x-sqrt3)+C$

I decomposed into basic fractions,

$(x^2-2x)/[(x^2-3)(x-3)(x-8)]$

= $(Ax+B)/(x^2-3)+C/(x-3)+D/(x-8)$

After expanding denominator,

$(Ax+B)(x-3)(x-8)+C(x^2-3)(x-8)+D(x^2-3)(x-3)=x^2-2x$

Set $x=3$ , $-30C=3$ , so $C=-1/10$

Set $x=8$ , $305D=48$ , so $D=48/305$

Set $x=0$ , $24B+24C+9D=0$ , so $B=-C-3/8*D=5/122$

Set $x=2$ , $14A+14B+14C+4D=-1$ , so $A=-7/122$

Hence,

$int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]$

= $-1/122int ((7x-5)*dx)/(x^2-3)-1/10int (dx)/(x-3)+5/122int (dx)/(x-8)$

= $5/122Ln(x-8)-1/10Ln(x-3)+C-1/122*int ((7x-5)*dx)/(x^2-3)$

Now I solved last integral,

$int ((7x-5)*dx)/(x^2-3)$

= $int ((7x-5)*dx)/[(x+sqrt3)(x-sqrt3)]$

= $(21-5sqrt3)/6int (dx)/(x-sqrt3)-(21+5sqrt3)/6int (dx)/(x+sqrt3)$

= $(21-5sqrt3)/6ln(x-sqrt3)-(21+5sqrt3)/6ln(x+sqrt3)$

Thus,

$int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]$

= $5/122Ln(x-8)+(21+5sqrt3)/732ln(x+sqrt3)-1/10Ln(x-3)-(21-5sqrt3)/6ln(x-sqrt3)+C$

How do you integrate f(x)=(x^2-2x)/((x^2-3)(x-3)(x-8))f(x)=(x^2-2x)/((x^2-3)(x-3)(x-8)) using partial fractions?