# Indefinite IntegralSolve the indefinite integral of 1/(x^2 - 4x)?

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### 2 Answers

You may also use the following alternative method, thus, you may complete the square to denominator, using the next formula, such that:

`a^2 - 2ab + b^2 = (a - b)^2`

Identifying `a^2 = x^2` and `2ab = 4x` , yields:

`{(a = x),(2xb = 4x):} => {(a = x),(b = 4/2):} => {(a = x),(b = 2):}`

Hence, completing the square, yields:

`(x^2 - 4x + 2^2) - 2^2 = (x - 2)^2 - 4`

Hence, you may substitute `(x - 2)^2 - 4` for `x^2 - 4x` such that:

`int 1/(x^2 - 4x) dx = int 1/((x - 2)^2 - 4)dx`

You should come up with the following substitution, such that:

`x - 2 = u => dx = du`

Changing the variable yields:

`int 1/((x - 2)^2 - 4)dx = int 1/(u^2 - 4) du`

You need to use the following formula such that:

`int 1/(x^2 - a^2) dx = 1/(2a)*ln|(x - a)/(x + a)| + c`

Reasoning by analogy yields:

`int 1/(u^2 - 4) du = 1/(2*2) ln|(u - 2)/(u + 2)| + c`

Substituting back `x - 2` for u yields:

`int 1/((x - 2)^2 - 4)dx = (1/4)ln|(x - 4)/(x)| + c`

**Hence, evaluating the given indefinite integral completing the square and using substitution, yields **`int 1/((x - 2)^2 - 4)dx = (1/4)ln|(x - 4)/(x)| + c.`

To determine the indefinite integral, we'll write the function as a sum or difference of elementary ratios.

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

1/x(x-4) = A/x + B/(x-4)

1 = A(x-4) + B(x)

We'll remove the brackets:

1 = Ax - 4A + Bx

We'll combine like terms:

1 = x(A+B) - 4A

A + B = 0

A = -B

-4A = 1

A = -1/4

B = 1/4

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

Int dx/x(x-4) = -Int dx/4x + Int dx/4(x-4)

Int dx/x(x-4) = -(1/4) (ln |x| - ln|x-4|) + C

Int dx/x(x-4) = -(1/4)ln |(x)/(x-4)| + C