Question

How do you integrate `(x^2-1)/(x^4-16)` using partial fractions?

Answer 1

`3/32ln|(x-2)/(x+2)|+5/16arc tan(x/2)+C`.

Explanation:

Let `I=int(x^2-1)/(x^4-16)dx`.

Here, the Integrand `(x^2-1)/(x^4-16)=(x^2-1)/((x^2-4)(x^2+4))`

`=(y-1)/((y-4)(y+4)), where, y=x^2` We split the Integrand using the

Partial Fractions. So, let,

`(y-1)/((y-4)(y+4))=A/(y-4)+B/(y+4), where, A,B in RR`.

To find, `A, &, B`, we use Heavyside's Cover-up Method :-

`A=[(y-1)/(y+4)]_(y=4)=(4-1)/(4+4)=3/8`.

`B=[(y-1)/(y-4)]_(y=-4)=(-4-1)/(-4-4)=5/8`. Hence,

`(y-1)/((y-4)(y+4))=(3/8)/(y-4)+(5/8)/(y+4)`. Letting, `y=x^2`,

`(x^2-1)/(x^4-16)=(3/8)/(x^2-4)+(5/8)/(x^2+4).` Therefore,

`I=3/8int1/(x^2-2^2)dx+5/8int1/(x^2+2^2)dx`

`=3/8*1/(2*2)ln|(x-2)/(x+2)|+5/8*1/2arc tan(x/2)`

`:. I=3/32ln|(x-2)/(x+2)|+5/16arc tan(x/2)+C`.

Enjoy Maths.!