# How do you integrate #1/(s^2 + s - 6)^2 # using partial fractions?

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To integrate (\frac{1}{(s^2 + s - 6)^2}) using partial fractions, first, factor the denominator (s^2 + s - 6):

(s^2 + s - 6 = (s + 3)(s - 2))

Now, express (\frac{1}{(s^2 + s - 6)^2}) as the sum of two partial fractions:

[\frac{1}{(s^2 + s - 6)^2} = \frac{A}{s + 3} + \frac{B}{s - 2}]

To find (A) and (B), we need to solve for them by equating the numerators of the original fraction and the decomposed fractions. Multiplying both sides by ((s^2 + s - 6)^2), we get:

[1 = A(s - 2) + B(s + 3)]

Expanding and equating coefficients, we get:

[1 = (A + B)s^2 + (B - 2A)s - 2A + 3B]

Equating coefficients:

[A + B = 0] [B - 2A = 1]

Solving these equations simultaneously, we find (A = -\frac{1}{5}) and (B = \frac{1}{5}).

Now, we have:

[\frac{1}{(s^2 + s - 6)^2} = \frac{-\frac{1}{5}}{s + 3} + \frac{\frac{1}{5}}{s - 2}]

Integrating each term separately, we get:

[\int \frac{1}{(s^2 + s - 6)^2} ds = -\frac{1}{5} \int \frac{1}{s + 3} ds + \frac{1}{5} \int \frac{1}{s - 2} ds]

[= -\frac{1}{5} \ln|s + 3| + \frac{1}{5} \ln|s - 2| + C]

Where (C) is the constant of integration.

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