Chapter 10.5, Problem 2QR

To determine

To find : whether the given expression is improper integral converges or diverges.

Answer to Problem 2QR

Hence, the limit is positive and ${{\displaystyle \int }}_1^{\infty }\frac{1}{x}dx$ diverges. Therefore, ${{\displaystyle \int }}_1^{\infty }\frac{x^2}{x^3+1}dx$ also diverges.

Explanation of Solution

Given information:

Given expression: ${{\displaystyle \int }}_1^{\infty }\frac{x^2}{x^3+1}dx$

Calculation:

The given integral is, ${{\displaystyle \int }}_1^{\infty }\frac{x^2}{x^3+1}dx$

We compare the terms of ${{\displaystyle \int }}_1^{\infty }\frac{x^2}{x^3+1}dx$ to terms of ${{\displaystyle \int }}_1^{\infty }\frac{1}{x}dx$ , Using Limit Comparing Test, $\underset{x\to \infty }{\lim }\frac{\frac{x^3+1}{1}}{\frac{1}{x}}=\underset{x\to \infty }{\lim }\frac{x^3}{x^3+1}=1$

Now, ${{\displaystyle \int }}_1^{\infty }\frac{1}{x}dx$ By $p-$ Series Test, ${{\displaystyle \int }}_1^{\infty }\frac{1}{x^p}dx$ converges if $p>1,$ and diverges if $p\le 1$

It is of the form of ${{\displaystyle \int }}_1^{\infty }\frac{1}{x^p}dx,$ where $p=1.$ Therefore, this series diverges, since $p=1.$

Hence, the limit is positive and ${{\displaystyle \int }}_1^{\infty }\frac{1}{x}dx$ diverges. Therefore, ${{\displaystyle \int }}_1^{\infty }\frac{x^2}{x^3+1}dx$ also diverges