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

To find int {1/(x^2-7 )}dx

Solution:

We know that 1/x^2-a^2 = (1/2a){1/(x-a) -1/(x+a)}, where a = sqrt7.

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

= (1/2a) {log (x-a) - log(x+a)}

= (1/2a){ log(x-a)/(x+a)}. Now put a = sqrt7

= 1/2sqrt7 { log(x-sqrt7)/(x+sqrt7)

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