# 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...

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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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