Find the area under the curve 1/(x^2-36) from the right of x=19?!

Hiiii. The question is
find the area under the curve 1/(x^2-36) from the right of x=19..

Thanks for all your help guys~

Integrate:

dx / (x^2 - 36) =>
A * dx / (x - 6) + B * dx / (x + 6)

A * (x + 6) + B * (x - 6) = 0x + 1
Ax + Bx + 6A - 6B = 0x + 1

Ax + Bx = 0x
A + B = 0

6A - 6B = 1
A - B = (1/6)

A + B + A - B = 0 + 1/6
2A = 1/6
A = 1/12
B = -1/12

(1/12) * dx / (x - 6) - (1/12) * dx / (x + 6) =>>
(1/12) * ln|x - 6| - (1/12) * ln|x + 6| + C =>
(1/12) * (ln|x - 6| - ln|x + 6|) + C =>
(1/12) * ln|(x - 6) / (x + 6)| + C

I suppose we're going from 19 to infinity?


(x - 6) / (x + 6) =>
(x + 6 - 12) / (x + 6) =>
(x + 6) / (x + 6) - 12 / (x + 6) =>
1 - 12 / (x + 6)

(1/12) * ln|1 - 12/(x + 6)| + C

From 19 to infinity:

(1/12) * ln|1 - 12/inf| - (1/12) * ln|1 - 12/25| =>
(1/12) * ln|1 - 0| - (1/12) * ln|13/25| =>
(1/12) * ln|1| + (1/12) * ln|25/13| =>
(1/12) * 0 + (1/12) * ln|25/13| =>
(1/12) * ln(25/13)