How do you find $\int \frac{x^{2} - 4}{(x + 3) (x^{2} + 1)} d x$ $int (x^2-4)/((x+3)(x^2+1))dx$ using partial fractions?

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Answer 1 · 2018-05-06T05:00:54

$\frac{1}{2} ln | (x + 3) | + \frac{1}{4} ln (x^{2} + 1) - \frac{3}{2} a r c tan x + C$ $1/2ln|(x+3)|+1/4ln(x^2+1)-3/2arc tanx+C$ .

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

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.!

How do you find int (x^2-4)/((x+3)(x^2+1))dx∫x2−4(x+3)(x2+1)dxint (x^2-4)/((x+3)(x^2+1))dx using partial fractions?