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

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