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

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UpStudy AI · Accepted Answer

The integral

$$
\int_{1}^{+\infty} \frac{x \sqrt{x} - 1}{x^{r} + x^{r} + 1} \, dx
$$

is an improper integral due to the upper limit being infinity. To evaluate it, we can analyze the behavior of the integrand as $$ x $$ approaches infinity and determine if the integral converges or diverges.

### Step 1: Simplify the integrand

First, let's rewrite the integrand:

$$
\frac{x \sqrt{x} - 1}{x^{r} + x^{r} + 1} = \frac{x^{3/2} - 1}{2x^{r} + 1}
$$

### Step 2: Analyze the behavior as $$ x \to +\infty $$

As $$ x $$ approaches infinity, we can approximate the integrand:

$$
\frac{x^{3/2} - 1}{2x^{r} + 1} \approx \frac{x^{3/2}}{2x^{r}} = \frac{1}{2} x^{3/2 - r}
$$

### Step 3: Determine convergence

The convergence of the integral depends on the exponent $$ 3/2 - r $$:

- If $$ 3/2 - r < -1 $$ (i.e., $$ r > 5/2 $$), the integral converges.
- If $$ 3/2 - r \geq -1 $$ (i.e., $$ r \leq 5/2 $$), the integral diverges.

### Conclusion

The integral converges if $$ r > \frac{5}{2} $$ and diverges if $$ r \leq \frac{5}{2} $$.

If you need to evaluate the integral for a specific value of $$ r $$ that satisfies the convergence condition, please provide that value.
The integral converges if $$ r > 2.5 $$ and diverges if $$ r \leq 2.5 $$.

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