Question

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

Accepted Answer

Let f(x) = 1/(x^2 -7)
Then.
F(x) = intg f(x)
         = intg 1/(x^2 -7)
But, we know that:
x^2 - 7 = (x-sqrt7)(x+sqrt7)
==> F(x) = intg 1/(x-sqrt7)(x+sqrt7)
Let us rewrite:
A/(x-sqrt7) + B/(x+sqrt7) = 1/(x-sqrt7)(x+sqrt7)
==> A(x+sqrt7) + B(x-sqrt7) = 1
==> Ax + Asqrt7 + Bx - Bsqrt7 = 1
==> (A+B)x + (A-B)sqrt7 = 1
==> A+B = 0==> A = -B
==> (A-B)sqrt7 = 1
==> (-2B)sqrt7 = 1
==> B = -1/2sqrt7
==> A = 1/2sqrt7
==> F(x) = intg 1/2sqrt7(x-sqrt7) - 1/2sqrt7(x+sqrt7)
               = (1/2sqrt7)*log (x-sqrt7) - (1/2sqrt7)*log (x+sqrt7)
              = (1/2sqrt7)*[log (x-sqrt7) - log (x+sqrt7)]
                = (1/2sqrt7)*log [(x-sqrt7)/(x+sqrt7)]

Answer

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)

Answer

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

Math

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

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