# Find the integral of #1/x^3+x#?

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To find the integral of ( \frac{1}{x^3 + x} ), we can use partial fraction decomposition. We first factor the denominator ( x^3 + x ) as ( x(x^2 + 1) ). Then, we write ( \frac{1}{x(x^2 + 1)} ) as ( \frac{A}{x} + \frac{Bx + C}{x^2 + 1} ). By finding the values of ( A ), ( B ), and ( C ), we can rewrite the expression as separate fractions and integrate each term individually. After integrating, we combine the results to find the final integral.

Starting with ( \frac{1}{x(x^2 + 1)} ), we get:

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

Multiplying both sides by ( x(x^2 + 1) ), we get:

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

Expanding and collecting like terms, we find:

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

Equating coefficients of like terms, we have:

[ A + B = 0 ] [ A + C = 0 ] [ B = 1 ]

Solving these equations, we find ( A = -B = -1 ) and ( C = -A = 1 ). Therefore, our expression becomes:

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

Integrating each term separately, we get:

[ \int \frac{-1}{x} ,dx = -\ln|x| + C_1 ]

[ \int \frac{x + 1}{x^2 + 1} ,dx = \frac{1}{2} \ln|x^2 + 1| + C_2 ]

Combining both results, we get the final integral:

[ \int \frac{1}{x^3 + x} ,dx = -\ln|x| + \frac{1}{2} \ln|x^2 + 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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