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!

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