Question

How do you find `int (x^2-4)/((x+3)(x^2+1))dx` using partial fractions?

Answer 1

`1/2ln|(x+3)|+1/4ln(x^2+1)-3/2arc tanx+C`.

Explanation:

Suppose that, `I=int(x^2-4)/{(x+3)(x^2+1)}dx`.

Let, `(x^2-4)/{(x+3)(x^2+1)}=A/(x+3)+(Bx+C)/(x^2+1), (A,B,C in RR)...(ast)`.

We will use Heavyside's Method to determine `A,B,C`.

`:. A=[(x^2-4)/(x^2+1)]_(x=-3)=(9-4)/(9+1)=1/2`.

Using `A` in `(ast)`, we have,

`(Bx+C)/(x^2+1)=(x^2-4)/{(x+3)(x^2+1)}-(1/2)/(x+3)`,

`={2(x^2-4)-(x^2+1)}/{2(x+3)(x^2+1)}, i.e.,`

`(Bx+C)/(x^2+1)=(x^2-9)/{2(x+3)(x^2+1)}={1/2(x-3)}/(x^2+1)`.

`:. B=1/2, and, C=-3/2`.

Altogether, we have,

`(x^2-4)/{(x+3)(x^2+1)}=(1/2)/(x+3)+(1/2x-3/2)/(x^2+1)`.

`:. I=1/2int1/(x+3)dx+1/2intx/(x^2+1)dx-3/2int1/(x^2+1)dx`,

`=1/2ln|(x+3)|+1/4int(2x)/(x^2+1)dx-3/2arc tanx`,

`=1/2ln|(x+3)|+1/4int{d/dx(x^2+1)}/(x^2+1)dx-3/2arc tanx`,

`rArr I=1/2ln|(x+3)|+1/4ln(x^2+1)-3/2arc tanx+C`.

Enjoy Maths.!