Find the partial fraction decomposition of the integrand. \( \int \frac{x^{2}+8 x-1}{x^{3}-x} d x=\int \square d x \)

To find the partial fraction decomposition of the integrand \(\frac{x^{2}+8x-1}{x^{3}-x}\), we start by factoring the denominator: \[ x^{3}-x = x(x^{2}-1) = x(x-1)(x+1) \] Next, we can set up the partial fraction decomposition as follows: \[ \frac{x^{2}+8x-1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} \] Multiplying through by the common denominator \(x(x-1)(x+1)\), we have: \[ x^{2}+8x-1 = A(x-1)(x+1) + B(x)(x+1) + C(x)(x-1) \] Expanding the terms on the right side and equating coefficients will give us the values for \(A\), \(B\), and \(C\). After solving, we can integrate each term separately: \[ \int \left( \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} \right) dx = A \ln|x| + B \ln|x-1| + C \ln|x+1| + C \] This results in the final integral form. Now, let’s break it down a bit more to ensure clarity in the solving process. To solve for \(A\), \(B\), and \(C\) in our equation \(x^{2} + 8x - 1 = A(x^{2}-1) + B(x^{2}+x) + C(x^{2}-x)\): 1. Substitute convenient values for \(x\) (e.g., 0, 1, -1). 2. Expand and collect like terms. 3. Create a system of equations based on the coefficients. After finding \(A\), \(B\), and \(C\): \[ \int \frac{x^{2}+8x-1}{x^{3}-x} \, dx = \int \left( A\frac{1}{x} + B\frac{1}{x-1} + C\frac{1}{x+1} \right) \, dx = A \ln|x| + B \ln|x-1| + C \ln|x+1| + C \] And that’s your integral solved through partial fractions!