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

1 Answer
Sep 21, 2016

#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.!