Integrate f(x) = 1/(x^2 - 4)

f(x) = 1/(x^2 - 4)

First
wewill implify using partial fractions:

==> 1/(x^2-
4) = 1/(x-2)(x+2)

==> A(x-2) + B(x+2) =
1/(x-2)(x+2)

==> A(x+2) + B(x-2) =
1

==> Ax + 2A + Bx - 2B =
1

==> (A+B)x + 2A-2B =
1

==> A+B = 0 ==> A =
-B

==>2(A-B)=
1

==>2(-B-B) =
1

==> -2B =
1/2

==>B =
-1/4

==>A =
1/4

==> 1/(x-2)(x+2) = 1/4(x-2) -
1/4(x+2)

Now let us
integrate:

intg f(x) = intg (1/4(x-2) - 1/4(x+2) ]
dx

             = (1/4) intg (1/(x-2) - 1/(x+2)]
dx

             = (1/4) intg (1/(x-2)dx - intg
(1/(x+2)dx

              = (1/4) * ln(x-2) - ln (x+2) +
C

              = (1/4) *ln(x-2)/(x+2) 
+C

==>intg (1/(x^2 -4)= (1/4) ln
(x-2)/(x+2) + C

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