Verify the result of indefinite integral of y = 1/(x^2 - 7)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Y = 1/(X^2 - 7)

INTG y = intg (1/(x^2 - 7))

Let us rewrite:

1/(x^2 -7) = 1/(x+sqrt7)(x-sqrt7)

==> 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 + (B-A)sqrt7 = 1

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

==: B-A = 1/sqrt7

==: 2B = 1/sqrt7

==> B = 1/2sqrt7

==> A = -1/2sqrt7

==> intg y = intg [ -1/sqrt7(x+sqrt7) + 1/2sqrt7*(x-sqrt7)

                   = (-1/2sqrt7)*[intg (1/(x+sqrt7) -1/(x-sqrt7)]

                    = (-1/2sqrt7)*intg(1/(x+sqrt7)- intg(1/(x-sqrt7)

                      =( -1/2sqrt6)*ln (x+sqrt7) - ln(x+sqrt7)

                      = (-1/2sqrt7)*ln [(x+sqrt7)/(x-sqrt7)]

         intg y   = ln[(x-sqrt7)/(x+sqrt7)]/ 2sqrt7

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To evaluate the indefinite integral, we calculate the denominator of the function, that is a difference of squares.

We'll re-write the function as a sum of 2 irreducible ratios:

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To determine integral of y = 1/(x^2-7).

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

We know that Integral of y = 1/(x^2-a^2) = Intetegral (1/2a) {dx/(x-a) - dx/(x+a)}

(1/2a) { ln(x-a) - ln(x+a)}

Integral y dx = (1/2a)  ln(x-a)/(x+a) + C.

Therefore  Integral dx/(x^2-7) = (1/2sqrt7) ln{ (x-sqrt7)/(x+sqt7) +C.

 

 

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