Integrate the function f(x)=1/(x^2-3).

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 quotients:

1/(x^2-3) =
1/(x-sqrt3)(x+sqrt3)

1/(x-sqrt3)(x+sqrt3) = A/(x-sqrt3) +
B/(x+sqrt3)

We'll multiply the first ratio from the right
side, by (x+sqrt3), and the second ratio, by (x-sqrt3).

1 =
A(x+sqrt3) + B(x-sqrt3)

We'll remove the brackets from the
right side:

1 = Ax + Asqrt3 + Bx -
Bsqrt3

We'll combine the like
terms:

1 = x(A+B) +
sqrt3(A-B)

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

A+B =
0

A = -B

sqrt3(A-B) =
1

We'll divide by sqrt3:

A-B =
1/sqrt3

A+A = 1/sqrt3

2A =
1/sqrt3

We'll divide by 2:

A =
1/2sqrt3

B = -1/2sqrt3

The
function 1/(x^2 - 3) = 1/2sqrt3(x-sqrt3) -
1/2sqrt3(x+sqrt3)

Int dx/(x^2 - 3) = (1/2sqrt3)*[Int
dx/(x-sqrt3) - Intdx/(x+sqrt3)]

We'll solve Int
dx/(x-sqrt3) using substitution technique:

We'll note
(x-sqrt3) = t

We'll differentiate both
sides:

dx = dt

 Int
dx/(x-sqrt3) = Int dt/t

Int dt/t = ln t + C = ln (x-sqrt3)
+ C

Intdx/(x+sqrt3) = ln (x+sqrt3) +
C

Int dx/(x^2 - 3) = (1/2sqrt3)*[ln (x-sqrt3)-ln (x+sqrt3)]
+ C

 We'll use the quotient property of the
logarithms:

Int dx/(x^2 - 3) = (1/2sqrt3)*[ln
(x-sqrt3)/(x+sqrt3)] + C