Find the area bounded by y= (x-12)/ (x^2 -4x -32), x=5, x=6 and y=0. For reference, previous answers to similar questions were very small, between 1 and 2.

Answer 1

desmos.com/calculator/8gkbyllvsi

Answer 2

y is positive in the x = [5,6] interval, so we can compute a single integral

A = ∫₅⁶y(x)dx = ∫₅⁶[4/3 1/(x+4) - 1/3 1/(x-8)] dx = 4/3 (ln|10| - ln|9|) - 1/3 l(ln|-2| - ln|-3|) =

= 4/3 ln(10/9) - 1/3 ln(2/3)

The trick is to decompose the ratio of polynomials in y(x) as the sum of simpler fractions, whose antiderivative is known. So, we set:

(x-12)/(x^2 -4x - 32) = (x-12)/[(x+4)(x-8)] = B/(x+4) + C/(x-8) = [(B+C)x + (4C - 8B)]/[(x+4)(x-8)].

You then need to solve. B + C = 1 and 4C - 8B = -12, as you equate the coefficients term by term