How do you integrate $\frac{x^{2} - 1}{x^{4} - 16}$ $(x^2-1)/(x^4-16)$ using partial fractions?

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Answer 1 · 2016-09-21T13:33:30

$\frac{3}{32} ln ∣ ∣ ∣ \frac{x - 2}{x + 2} ∣ ∣ ∣ + \frac{5}{16} a r c tan (\frac{x}{2}) + C$ $3/32ln|(x-2)/(x+2)|+5/16arc tan(x/2)+C$ .

Let $I = \int \frac{x^{2} - 1}{x^{4} - 16} d x$ $I=int(x^2-1)/(x^4-16)dx$ .

Here, the Integrand $\frac{x^{2} - 1}{x^{4} - 16} = \frac{x^{2} - 1}{(x^{2} - 4) (x^{2} + 4)}$ $(x^2-1)/(x^4-16)=(x^2-1)/((x^2-4)(x^2+4))$

$= \frac{y - 1}{(y - 4) (y + 4)}, w h e r e, y = x^{2}$ $=(y-1)/((y-4)(y+4)), where, y=x^2$ 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.!

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