# Evaluate # int \ 1/(x^2-2x) \ dx #?

# int \ 1/(x^2-2x) \ dx = 1/2 ln |(A(x-2))/x| #

We want to find:

If we examine the integrand we can decompose into partial fractions, as follows:

And so:

Put:

Therefore:

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To evaluate the integral (\int \frac{1}{x^2 - 2x} , dx), we first factor the denominator:

[x^2 - 2x = x(x - 2)]

Now, we use partial fraction decomposition to break down the fraction:

[\frac{1}{x(x - 2)} = \frac{A}{x} + \frac{B}{x - 2}]

Multiplying both sides by (x(x - 2)), we get:

[1 = A(x - 2) + Bx]

Expanding and collecting like terms:

[1 = (A + B)x - 2A]

This equation must hold for all values of (x), so the coefficients must be equal:

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

Solving this system of equations, we find (A = -\frac{1}{2}) and (B = \frac{1}{2}).

Now, we can rewrite the integral as:

[\int \left(-\frac{1}{2x} + \frac{1}{2(x - 2)}\right) , dx]

Which integrates to:

[-\frac{1}{2} \ln |x| + \frac{1}{2} \ln |x - 2| + C]

Where (C) is the constant of integration.

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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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