\( \int _ { 1 } ^ { + \infty } \frac { x \sqrt { x } - 1 } { x ^ { r } + x ^ { r } + 1 } d x \)

To evaluate the integral \( \int _ { 1 } ^ { + \infty } \frac { x \sqrt { x } - 1 } { x ^ { r } + x ^ { r } + 1 } d x \), we begin by simplifying the expression in the integrand. Note that \( x \sqrt{x} - 1 = x^{3/2} - 1 \). As \( x \to \infty \), the dominant term in the denominator, \( x^r + x^r + 1 \), behaves like \( 2x^r \). This allows us to rewrite the integrand for large \( x \): \[ \frac{x^{3/2} - 1}{x^r + x^r + 1} \approx \frac{x^{3/2}}{2x^r} = \frac{1}{2} x^{3/2 - r} \quad \text{for large } x. \] To determine convergence of the integral, we analyze the behavior based on the value of \( r \): - If \( r < \frac{3}{2} \), then \( 3/2 - r > 0 \) implies the integral diverges because the integrand behaves like \( \frac{1}{2} x^{3/2 - r} \). - If \( r = \frac{3}{2} \), the integrand behaves like \( \frac{1}{2} \), leading the integral to diverge as well. - If \( r > \frac{3}{2} \), we have \( 3/2 - r < 0 \), and thus, the integral converges since the integrand behaves like \( \frac{1}{2} x^{3/2 - r} \) which goes to 0. Now, we can write: \[ \int _{1}^{+\infty} \frac{x \sqrt{x} - 1}{x^r + x^r + 1} \, dx \quad \text{converges if } r > \frac{3}{2}. \] We conclude the assessment by evaluating at \( r = \frac{3}{2} \) which leads to divergence and finding the points of convergence at \( r > \frac{3}{2} \). For \( r < \frac{3}{2} \), the result diverges indicating that special care must be taken while analyzing the integral behavior based on \( r \)'s value.