We use,
(a-b)^2=a^2-2ab+b^2(a−b)2=a2−2ab+b2
and a^2-b^2=(a+b)(a-b)a2−b2=(a+b)(a−b)
to factorise the denominator
x^4-8x^2+16=(x^2-4)^2x4−8x2+16=(x2−4)2
=(x+2)^2(x-2)^2=(x+2)2(x−2)2
So,
(5x^3+2x^2-12x-8)/(x^4-8x^2+16)=(5x^3+2x^2-12x-8)/((x+2)^2(x-2)^2)5x3+2x2−12x−8x4−8x2+16=5x3+2x2−12x−8(x+2)2(x−2)2
=A/(x+2)^2+B/(x+2)+C/(x-2)^2+D/(x-2)=A(x+2)2+Bx+2+C(x−2)2+Dx−2
=(A(x-2)^2+B(x+2)(x-2)^2+C(x+2)^2+D(x-2)(x+2)^2)/((x+2)^2(x-2)^2)=A(x−2)2+B(x+2)(x−2)2+C(x+2)2+D(x−2)(x+2)2(x+2)2(x−2)2
So,
5x^3+2x^2-12x-8=A(x-2)^2+B(x+2)(x-2)^2+C(x+2)^2+D(x-2)(x+2)^25x3+2x2−12x−8=A(x−2)2+B(x+2)(x−2)2+C(x+2)2+D(x−2)(x+2)2
Let x=2x=2, =>⇒, 16=16C16=16C, =>⇒, C=1C=1
Let x=-2x=−2, =>⇒, -16=16A−16=16A, =>⇒, A=-1A=−1
Coefficients of x^3x3, 5=B+D5=B+D
And -8=4A+8B+4C-8D−8=4A+8B+4C−8D
-8=-4+8B+4-8D−8=−4+8B+4−8D
-8=8B-8D−8=8B−8D
B-D=-1B−D=−1
So, B=2B=2 and D=3D=3
int((5x^3+2x^2-12x-8)dx)/(x^4-8x^2+16)∫(5x3+2x2−12x−8)dxx4−8x2+16
==int(-1dx)/(x+2)^2+int(2dx)/(x+2)+int(1dx)/(x-2)^2+int(3dx)/(x-2)==∫−1dx(x+2)2+∫2dxx+2+∫1dx(x−2)2+∫3dxx−2
=1/(x+2)+2ln(∣x+2∣)-1/(x-2)+3ln(∣x-2∣)+C=1x+2+2ln(∣x+2∣)−1x−2+3ln(∣x−2∣)+C
