# How do you evaluate the integral #int (x^5+3x^2+1)/(x^4-1)dx#?

The answer is

We need

As the degree of the numerator is greater than the degree of the denominator, perform a long division first.

Perform a decomposition into partial fractions

The denominators are the same , compare the numerators

Therefore,

So,

Finally,

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To evaluate the integral (\int \frac{x^5+3x^2+1}{x^4-1}dx), we first perform polynomial long division to simplify the integrand. The result of the division is (x + \frac{3x^2}{x^4-1}).

Next, we decompose (\frac{3x^2}{x^4-1}) into partial fractions.

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

Solving for (A), (B), (C), and (D) yields (A = \frac{3}{2}), (B = -\frac{3}{2}), (C = 0), and (D = 0).

So, the integral becomes:

[\int \left( x + \frac{3}{2(x-1)} - \frac{3}{2(x+1)} \right) dx]

Integrating term by term, we get:

[\frac{x^2}{2} + \frac{3}{2}\ln|x-1| - \frac{3}{2}\ln|x+1| + C]

where (C) is the constant of integration.

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