# Evaluate the indefinite integral of y= 1/(x^2 - 11)

william1941 | College Teacher | (Level 3) Valedictorian

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We have to find the integral of y= 1/(x^2 - 11).

Now, we know that the integral of 1/x is ln x.

Now 1/(x^2 - 11)

= 1 / [(x- sqrt 11) ( x+ sqrt 11)]

= (1/ 2*sqrt 11)* [{1/ (x- sqrt 11)} - {1/ ( x+ sqrt 11)}]

Now integrating this we get:

(1/ 2*sqrt 11)* [ ln (x- sqrt 11) - ln ( x+ sqrt 11)]

= (1/ 2*sqrt 11)* ln [(x- sqrt 11) / ( x+ sqrt 11)]

The required result is :

(1/ 2*sqrt 11)* ln [(x- sqrt 11) / ( x+ sqrt 11)] + C

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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 ratios:

1/(x^2-11) = 1/(x-sqrt11)(x+sqrt11)

1/(x-sqrt11)(x+sqrt11) = A/(x-sqrt11) + B/(x+sqrt11)

We'll multiply the first ratio from the right side, by (x+sqrt11), and the second ratio, by (x-sqrt11).

1 = A(x+sqrt11) + B(x-sqrt11)

We'll remove the brackets from the right side:

1 = Ax + Asqrt11+ Bx - Bsqrt11

We'll combine the like terms:

1 = x(A+B) + sqrt11(A-B)

For the equality to hold, the like terms from both sides have to be equal:

A+B = 0

A = -B

sqrt11(A-B) = 1

We'll divide by sqrt11:

A-B = 1/sqrt11

A+A = 1/sqrt11

2A = 1/sqrt11

We'll divide by 2:

A = 1/2sqrt11

B = -1/2sqrt11

The function 1/(x^2-11) = 1/2sqrt11(x+sqrt11) - 1/2sqrt11(x-sqrt11)

Int dx/(x^2-11) = (1/2sqrt11)*[Int dx/(x-sqrt11) - Intdx/(x+sqrt11)]

We'll solve Int dx/(x-sqrt11) using substitution technique:

We'll note (x-sqrt11) = t

We'll differentiate both sides:

dx = dt

Int dx/(x-sqrt11) = Int dt/t

Int dt/t = ln t + C = ln (x+sqrt11) + C

Intdx/(x+sqrt11) = ln (x+sqrt11) + C

Int dx/(x^2 - 11) = (1/2sqrt11)*[ln (x-sqrt11)-ln (x+sqrt11)] + C

We'll use the quotient property of the logarithms:

Int dx/(x^2 - 11) = (1/2sqrt11)*[ln (x-sqrt11)/(x+sqrt11)] + C

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To integrate y = 1/(x^2=11)

Solution:

We shall slplit 1/(x^2-a^2) in to partial fractions, where a = (11)^(1/2)

1/(x-a^2) = 1/(x-a) -1/1(x+a)

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

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

Integral  dx/(x^2-a^2) = (1/2a) ln[(x-a)/(x+a)] + Constant of integration.

Now put = a= 11Integral dx/(x^2-11) = (1/2sqrt11)ln[(x-sqrt11)/(x-sqrt11)] +C