# How do you evaluate the integral #int (2x-3)/((x-1)(x+4))#?

Use partial fractions decomposition:

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To evaluate the integral (\int \frac{2x-3}{(x-1)(x+4)} , dx), we first use partial fraction decomposition to express the integrand as the sum of two simpler fractions. The decomposition is:

(\frac{2x-3}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4})

By equating numerators, we find:

(2x - 3 = A(x+4) + B(x-1))

Expanding and equating coefficients, we get:

(2x - 3 = (A + B)x + (4A - B))

This gives us a system of equations:

- (A + B = 2)
- (4A - B = -3)

Solving this system, we find (A = \frac{5}{5}) and (B = \frac{-3}{5}). Now we integrate:

(\int \frac{5}{x-1} - \frac{3}{x+4} , dx)

Using the fact that (\int \frac{1}{x} , dx = \ln|x|), we integrate term by term:

(5\ln|x-1| - 3\ln|x+4| + C)

Where (C) is the constant of integration. Thus, the integral of (\frac{2x-3}{(x-1)(x+4)}) is (5\ln|x-1| - 3\ln|x+4| + C).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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