How do you evaluate the integral #int dx/(x^4-16)#?

Question

Emilia Aguilar · Accepted Answer

# 1/32ln|(x-2)/(x+2)|-1/16arc tan(x/2)+C#.

Let, #I=intdx/(x^4-16)=intdx/{(x^2+4)(x^2-4)}# #=1/8int8/{(x^2+4)(x^2-4)}dx#, #=1/8int{(x^2+4)-(x^2-4)}/{(x^2+4)(x^2-4)}dx#, #=1/8int{(x^2+4)/{(x^2+4)(x^2-4)}-(x^2-4)/{(x^2+4)(x^2-4)}}dx#, #=1/8int{1/(x^2-4)-1/(x^2+4)}dx#, #=1/8{int1/(x^2-2^2)-int1/(x^2+2^2)dx}#, #=1/8{1/(2xx2)ln|(x-2)/(x+2)|-1/2arc tan(x/2)}#, #rArr I=1/32ln|(x-2)/(x+2)|-1/16arc tan(x/2)+C#.

Luke Alvarez · Answer

undefined To evaluate the integral ∫ dx/(x^4-16), we can first factor the denominator as the difference of squares: x^4 - 16 = (x^2)^2 - (4)^2 = (x^2 - 4)(x^2 + 4). We can then use partial fraction decomposition to break down the integrand into simpler fractions. The decomposition will be of the form A/(x-2) + B/(x+2) + C/(x^2 + 4), where A, B, and C are constants to be determined. Once we find the values of A, B, and C, we can integrate each term separately. The integral of A/(x-2) and B/(x+2) are straightforward to evaluate using the natural logarithm function. For the term C/(x^2 + 4), we use the arctangent function. Finally, we combine the integrals of each term to obtain the overall result.