# How do you find #int (x^3-x-1)/((x^4-1)x) dx# using partial fractions?

I decomposed integrand into basic fractions,

After expanding denominators,

Hence,

Hence,

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To find ( \int \frac{x^3 - x - 1}{(x^4 - 1)x} dx ) using partial fractions, we first factor the denominator ( x^4 - 1 ) as ( (x^2 + 1)(x + 1)(x - 1) ). Then, we express the given fraction as a sum of simpler fractions with denominators of ( x^2 + 1 ), ( x + 1 ), and ( x - 1 ). The general form of the partial fraction decomposition is:

[ \frac{x^3 - x - 1}{(x^4 - 1)x} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1} + \frac{Ex + F}{x - 1} + \frac{Gx + H}{x + 1} ]

Multiplying both sides by ( (x^4 - 1)x ) to clear the fractions, we get:

[ x^3 - x - 1 = A(x^3 - x) + B(x^2 - 1) + (Cx + D)(x^3 - x) + (Ex + F)(x^2 + 1) + (Gx + H)(x^2 + 1) ]

Now, we can equate coefficients of like terms on both sides to solve for ( A ), ( B ), ( C ), ( D ), ( E ), ( F ), ( G ), and ( H ). Once we find these values, we can integrate each term individually to find the integral of the original function.

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