What is the indefnite integral of 1/(x^2

Wednesday, April 13, 2011

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

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