How do you use partial fraction decomposition to decompose the fraction to integrate #(6x)/((x-4) (x+4))#?

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Answer 1 · 2018-03-28T21:16:49

The integral is #3ln|x^2 - 16| + C#

We have:

#A/(x- 4) + B/(x +4) = (6x)/((x -4)(x + 4))#

#A(x + 4) + B(x -4) = 6x#

#Ax+ Bx + 4A- 4B = 6x#

Thus #{(A + B = 6), (4A - 4B = 0):}#

Solving we get

#2A = 6#

#A = 3#

#B = 3#

Therefore the integral becomes #int 3/(x -4) + 3/(x +4) dx#. This is readily integrated as #3ln|x- 4| + 3ln|x + 4| = 3ln|x^2 - 16| + C##

Hopefully this helps!