# Calculate the indefinite integral of y = x^2/(x^2-9)?

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### 1 Answer

We notice that if we'll subtract 9 to numerator, we'll get the same expression with the one from denominator. But if we've subtracted 9, we'll have to add 9, to hold the given function.

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

Now, we'll calculate the indefinite integral:

Int f(x)dx = Int x^2dx/(x^2-9)

Int x^2dx/(x^2-9) = Int(x^2-9)dx/(x^2-9) + Int 9dx/(x^2-9)

But Int(x^2-9)dx/(x^2-9) = Int dx

Int x^2dx/(x^2-9) = Int dx + Int 9dx/(x^2-9)

We'll calculate Int 9dx/(x^2-9).

Since 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^2-9) = 1/(x-3)(x+3)

1/(x-3)(x+3) = A/(x-3) + B/(x+3)

We'll multiply the first ratio from the right side, by (x+3), and the second ratio, by (x-3).

1 = A(x+3) + B(x-3)

We'll remove the brackets from the right side:

1 = Ax+ 3A+ Bx - 3B

We'll combine the like terms:

1 = x(A+B) + 3(A-B)

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

A+B = 0

A = -B

3(A-B) = 1

We'll divide by 3:

A-B = 1/3

A+A = 1/3

2A = 1/3

We'll divide by 2:

A = 1/6

B = -1/6

The function 1/(x^2-9) = 1/6(x-3) - 1/6(x+3)

Int dx/(x^2-9) = (1/6)*[Int dx/(x-3) - Intdx/(x+3)]

We'll solve Int dx/(x-3) using substitution technique:

We'll note (x-3) = t

We'll differentiate both sides:

dx = dt

Int dx/(x-3) = Int dt/t

Int dt/t = ln t + C = ln (x-3) + C

Intdx/(x+3) = ln (x+3) + C

Int dx/(x^2 - 9) = (1/6)*[ln (x-3)-ln (x+3)] + C

We'll use the quotient property of the logarithms:

**Int dx/(x^2 - 9) = (1/6)*[ln (x-3)/(x+3)] + C**

Int x^2dx/(x^2-9) = Int dx + Int 9dx/(x^2-9)

**Int x^2dx/(x^2-9) = x + (9/6)*[ln (x-3)/(x+3)] + C**