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

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### 4 Answers

f(x) = 1/(x^2-3)

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

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

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

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

==> A+B = 1 ==> A= -B

==> (A-B)sqrt3 = 1

==? A-B = 1/sqrt3

==> -2b = 1/sqrt3

==> B = -1/2sqrt3

==> A = 1/2sqrt3

==> f(x) = 1/2sqrt3(x+sqrt3) - 1/2sqrt3(x-sqrt3)

==> intg f(x) = intg 1/2sqrt3(x+sqrt3) - intg 1/2sqrt3(x-sqrt3)

= (1/2sqrt3)(ln x+sqrt3 - ln (x-sqrt3)

= (1/2sqrt3)*ln (x+sqrt3/(x-sqrt3)

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**

Here we use the relation:

Int (1/x) dx = ln x

For the given function f(x) = 1/ (x^2 -3)

=> 1/ ( x - sqrt 3)( x+ sqrt 3)

=> [1/(- 2sqrt 3)]* [1/ (x+ sqrt 3)- 1/(x- sqrt 3)]

=> [1/(- 2sqrt 3)]* [1/ (x+ sqrt 3) - [1/(- 2sqrt 3)]* [1/ (x- sqrt 3)]

Int [1/(- 2sqrt 3)]* [1/ (x+ sqrt 3) - [1/(- 2sqrt 3)]* [1/ (x- sqrt 3)]

= [1/(- 2sqrt 3)]* ln (x+ sqrt 3) - [1/(- 2sqrt 3)]* ln (x- sqrt 3)

= [1/(- 2sqrt 3)]* ln [(x+ sqrt 3)/ (x- sqrt 3)]

**Therefore the result is [1/(- 2sqrt 3)]* ln [(x+ sqrt 3)/ (x- sqrt 3)]**

To find integral of f(x) = 1/(x^2-3)

f(x) = 1/x^2-3= 1/(x^2-a^2), where a^2= 3.

Therefore f(x) = (1/2a) {1/(x-a) -1/(x+a)}

Therefore Integral f(x) dx = inttegral (1/2a){1/(x-a) - 1/(x+a)} dx

= (1/2a) Integral {dx/(x-a) - dx/(x+a)}

= (1/2a){log(x-a) = log(x+a)} .

= (1/2a) log {(x-a)/(x+a)}

= (1/2sqrt3){log{(x-sqrt3)/(x+sqrt3)}