# How do you integrate #int(x+1)/((x^2-3)(x-1))# using partial fractions?

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To integrate (\frac{{x + 1}}{{(x^2 - 3)(x - 1)}}) using partial fractions, you first express it as:

[ \frac{{x + 1}}{{(x^2 - 3)(x - 1)}} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 - 3} ]

Solve for (A), (B), and (C) by equating numerators:

[ x + 1 = A(x^2 - 3) + (Bx + C)(x - 1) ]

Expand and equate coefficients of like terms:

- Equate constants: (A - C = 1)
- Equate coefficients of (x): (B - A = 0)
- Equate coefficients of (x^2): (A = 0)

Solve for (A), (B), and (C):

- (A = 0)
- (B = 0)
- (C = -1)

Now integrate using the partial fractions:

[ \int \left( \frac{A}{x - 1} + \frac{Bx + C}{x^2 - 3} \right) , dx = \int \left( \frac{-1}{x - 1} + \frac{0x - 1}{x^2 - 3} \right) , dx ]

The integral of (\frac{-1}{x - 1}) is (-\ln|x - 1|), and the integral of (\frac{-1}{x^2 - 3}) can be expressed in terms of inverse trigonometric functions.

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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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