Calculate the indefinite integral of 1/(x^2-7) .

Because the denominator of the function is a difference of
squares, we'll re-write the function as a sum of 2 irreducible
quotients:

1/(x^2-7) =
1/(x-sqrt7)(x+sqrt7)

1/(x-sqrt7)(x+sqrt7) = A/(x-sqrt7) +
B/(x+sqrt7)

We'll multiply the first ratio from the right
side, by (x+sqrt7), and the second ratio, by (x-sqrt7).

1 =
A(x+sqrt7) + B(x-sqrt7)

We'll remove the brackets from the
right side:

1 = Ax + Asqrt7 + Bx -
Bsqrt7

We'll combine the like
terms:

1 = x(A+B) +
sqrt7(A-B)

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

A+B =
0

A = -B

sqrt7(A-B) =
1

We'll divide by sqrt7:

A-B =
1/sqrt7

A+A = 1/sqrt7

2A =
1/sqrt7

We'll divide by 2:

A =
1/2sqrt7

B = -1/2sqrt7

The
function 1/(x^2 - 7) = 1/2sqrt7(x-sqrt7) -
1/2sqrt7(x+sqrt7)

Int dx/(x^2 - 7) = (1/2sqrt7)*[Int
dx/(x-sqrt7) - Intdx/(x+sqrt7)]

We'll solve Int
dx/(x-sqrt7) using substitution technique:

We'll note
(x-sqrt7) = t

We'll differentiate both
sides:

dx = dt

 Int
dx/(x-sqrt7) = Int dt/t

Int dt/t = ln t + C = ln (x-sqrt7)
+ C

Intdx/(x+sqrt7) = ln (x+sqrt7) +
C

Int dx/(x^2 - 7) = (1/2sqrt7)*[ln (x-sqrt7)-ln (x+sqrt7)]
+ C

 We'll use the quotient property of the
logarithms:

Int dx/(x^2 - 7) = (1/2sqrt7)*[ln
(x-sqrt7)/(x+sqrt7)] + C