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

You try to split the rational function into a sum that will be really easy to integrate.

Partial fraction decomposition allows you to do that :

In order to find them, you have to multiply both sides by one of the polynomials at the left of the equality. I show one example to you, the other coefficient is to be found the same way.

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To find the integral of (\frac{x+1}{x(x^2-1)}) using partial fractions, we first decompose the fraction into partial fractions:

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

Next, we clear the denominators by multiplying both sides by (x(x^2-1)):

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

Expanding and collecting like terms:

[ x+1 = Ax^2 - A + Bx^2 - Bx + Cx^2 + Cx ]

Grouping terms with the same powers of (x):

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

This equation yields the following system of equations:

[ \begin{align*} A + B + C &= 0 \ -B + C &= 1 \ -A &= 1 \end{align*} ]

Solving this system, we find (A = -1), (B = 0), and (C = 1).

Substitute these values back into the partial fraction decomposition:

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

Now, integrate each term separately:

[ \int \frac{-1}{x} dx + \int \frac{1}{x-1} dx ]

This gives:

[ -\ln|x| + \ln|x-1| + 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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