# Calculate the integral of y=(x^2+5)/(x^2-4)

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### 3 Answers

Here we use the relation:

Int (1/x) dx = ln x

y= ( x^2+ 5)/ (x^2-4)

= ( x^2- 4 + 9) / (x^2-4)

= (x^2-4) / (x^2-4) + 9/ (x^2-4)

=1+ 9/ (x^2-4)

= 1 + 9/ (x-2 )*( x+2)

= 1+ (9/-4)*[ 1/(x+2) - 1 / (x-2) ]

= 1 - (9/4)(1/(x+2)) + (9/4)(1/(x-2))

Int [1 - (9/4)(1/(x+2)) + (9/4)(1/(x-2))] dx

= x - (9/4) ln ( x+2) + (9/4) ln (x-2) + C

**The required result is x - (9/4) ln ( x+2) + (9/4) ln (x-2) + C**

To find Integral (x^2+5)/(x^2-4) dx.

Solution:

Int (x^2+5)/(x^2-4 dx = Int{(x^2-4) + 9}/(x^2-4)} dx =

int {(x^2-4)/x^2-4) +9/(x^2-4)} dx = Int 1dx + Int 9dx/(x^2-4)= x+9Int (1/2){ 1/x-2 - 1/x+2)} dx = x + 9(1/2) log (x-2) -log(x+2)}+Constant

Int (x^2+5)/(x^2-4) dx = x+ log(x-2)/(x+2) +Const.

We'll re-write the function in this way:

(x^2-4+9)/(x^2-4) = (x^2-4)/(x^2-4) + (9)/(x^2-4)

y = 1 + 9/(x^2-4)

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^2-4) = 1/(x-2)(x+2)

1/(x-2)(x+2) = A/(x-2) + B/(x+2)

We'll multiply the first ratio from the right side, by (x+2), and the second ratio, by (x-2).

1 = A(x+2) + B(x-2)

We'll remove the brackets from the right side:

1 = Ax + 2A + Bx - 2B

We'll combine the like terms:

1 = x(A+B) + 2(A-B)

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

A+B = 0

A = -B

2(A-B) = 1

We'll divide by 2:

A-B = 1/2

A+A = 1/2

2A = 1/2

We'll divide by 2:

A = 1/4

B = -1/4

The function 1/(x^2 - 4) = 1/4(x-2) - 1/4(x+2)

Int dx/(x^2 - 4) = (1/4)*[Int dx/(x-2) - Intdx/(x+2)]

We'll solve Int dx/(x-2) using substitution technique:

We'll note (x-2) = t

We'll differentiate both sides:

dx = dt

Int dx/(x-2) = Int dt/t

Int dt/t = ln t + C = ln (x-2) + C

Intdx/(x+2) = ln (x+2) + C

Int dx/(x^2 - 4) = (1/4)*[ln (x-2)-ln (x+2)] + C

We'll use the quotient property of the logarithms:

**Int dx/(x^2 - 4) = (1/4)*[ln (x-2)/(x+2)] + C**