# What is the indefnite integral of 1/(x^2 - 4x)?

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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**