Evaluate the indefinite integral of y= 1/(x^2 - 11)

We notice that the denominator of the function is a
difference of squares.

We'll re-write the function as a sum
of 2 irreducible ratios:

1/(x^2-11) =
1/(x-sqrt11)(x+sqrt11)

1/(x-sqrt11)(x+sqrt11) =
A/(x-sqrt11) + B/(x+sqrt11)

We'll multiply the first ratio
from the right side, by (x+sqrt11), and the second ratio, by
(x-sqrt11).

1 = A(x+sqrt11) +
B(x-sqrt11)

We'll remove the brackets from the right
side:

1 = Ax + Asqrt11+ Bx -
Bsqrt11

We'll combine the like
terms:

1 = x(A+B) +
sqrt11(A-B)

For the equality to hold, the like terms from
both sides have to be equal:

A+B =
0

A = -B

sqrt11(A-B) =
1

We'll divide by sqrt11:

A-B
= 1/sqrt11

A+A = 1/sqrt11

2A =
1/sqrt11

We'll divide by 2:

A
= 1/2sqrt11

B = -1/2sqrt11

The
function 1/(x^2-11) = 1/2sqrt11(x+sqrt11) -
1/2sqrt11(x-sqrt11)

Int dx/(x^2-11) = (1/2sqrt11)*[Int
dx/(x-sqrt11) - Intdx/(x+sqrt11)]

We'll solve Int
dx/(x-sqrt11) using substitution technique:

We'll note
(x-sqrt11) = t

We'll differentiate both
sides:

dx = dt

 Int
dx/(x-sqrt11) = Int dt/t

Int dt/t = ln t + C = ln
(x+sqrt11) + C

Intdx/(x+sqrt11) = ln (x+sqrt11) +
C

Int dx/(x^2 - 11) = (1/2sqrt11)*[ln (x-sqrt11)-ln
(x+sqrt11)] + C

 We'll use the quotient property of the
logarithms:

Int dx/(x^2 - 11) =
(1/2sqrt11)*[ln (x-sqrt11)/(x+sqrt11)] + C