# How do you use partial fractions to find the integral #int (x^2-1)/(x^3+x)dx#?

Start by factoring the denominator.

Now write the partial fraction decomposition.

We now write a system of equations.

Hopefully this helps!

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To integrate ( \frac{x^2 - 1}{x^3 + x} ) using partial fractions, we first factor the denominator:

[ x^3 + x = x(x^2 + 1) ]

Since the denominator does not factor further over the real numbers, we can use the following partial fractions decomposition:

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

Multiplying both sides by ( x^3 + x ), we get:

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

Comparing coefficients, we find:

For ( x^2 ): ( A + B = 1 )

For ( x ): ( C = 0 )

For the constant term: ( A = -1 )

Therefore, we have:

[ \frac{x^2 - 1}{x^3 + x} = \frac{-1}{x} + \frac{1}{x^2 + 1} ]

Now, we can integrate term by term:

[ \int \frac{x^2 - 1}{x^3 + x} dx = \int \left( \frac{-1}{x} + \frac{1}{x^2 + 1} \right) dx ] [ = -\ln|x| + \arctan(x) + C ]

So, the integral of ( \frac{x^2 - 1}{x^3 + x} ) is ( -\ln|x| + \arctan(x) + 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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