14 Multiple Choice 1 point Evaluate the integral: \( \int_{2}^{\infty} \frac{x^{2}}{\left(x^{3}-1\right)^{4}} d x \)

To evaluate the integral \( \int_{2}^{\infty} \frac{x^{2}}{(x^{3}-1)^{4}} \, dx \), we first consider the behavior of the integrand as \( x \to \infty \). For large \( x \), the term \( x^3 - 1 \) behaves like \( x^3 \), so we can approximate the integrand: \[ \frac{x^{2}}{(x^{3}-1)^{4}} \approx \frac{x^{2}}{(x^{3})^{4}} = \frac{x^{2}}{x^{12}} = \frac{1}{x^{10}} \] Now we know that \( \int \frac{1}{x^{10}} \, dx \) converges at infinity. Thus, our original integral converges. For the actual calculation, we perform a substitution. Let \( u = x^3 - 1 \), then \( du = 3x^2 \, dx \) or \( dx = \frac{du}{3x^2} \). When \( x = 2 \), we find: \[ u = 2^3 - 1 = 8 - 1 = 7 \] As \( x \to \infty \), \( u \to \infty \). Therefore the bounds change from 2 to \( \infty \) in terms of \( u \), the integral becomes: \[ \int_{7}^{\infty} \frac{x^2}{(u)^{4}} \cdot \frac{du}{3x^{2}} = \frac{1}{3} \int_{7}^{\infty} \frac{1}{u^{4}} \, du \] Now, we compute \( \int \frac{1}{u^4} \, du \): \[ \int \frac{1}{u^{4}} \, du = -\frac{1}{3u^{3}} + C \] Thus, we evaluate the definite integral: \[ \frac{1}{3} \left[-\frac{1}{3u^{3}}\right]_{7}^{\infty} = \frac{1}{3} \left( 0 + \frac{1}{3 \cdot 7^{3}} \right) = \frac{1}{3 \cdot 3 \cdot 343} = \frac{1}{1029} \] Finally, the result of the integral \( \int_{2}^{\infty} \frac{x^{2}}{(x^{3}-1)^{4}} \, dx \) is: \[ \boxed{\frac{1}{1029}} \]