Question

What, if anything, does the divergence of ${\int }_1^{\infty }\frac{1}{x^2}dx$and the comparison test tell you about the convergence or divergence of ${\int }_1^{\infty }\frac{1}{x^2+1}dx$and why?

Short answer

The convergence of${\int }_1^{\infty }\frac{1}{x^2}dx$ and comparison test are useful to tell the convergence of ${\int }_1^{\infty }\frac{1}{x^2+1}dx$.

Step-by-step solution

Step 1. Given information.   

We are given two integrals,

${\int }_1^{\infty }\frac{1}{x^2}dx$and

${\int }_1^{\infty }\frac{1}{x^2+1}dx$

Step 2. Comparing the integrals. 

Finding the relation between $\frac{1}{x^2}$and $\frac{1}{x^2+1}$in the interval $[1,\infty )$.

$x^2<x^2+1$for all $x\in [1,\infty )$

$\frac{1}{x^2}>\frac{1}{x^2+1}$for all $x\in [1,\infty )$

That is, $\frac{1}{x^2+1}<\frac{1}{x^2}$for all $x\in [1,\infty )$

Apply comparison test for improper integrals to the continuous functions, we get

Since ${\int }_1^{\infty }\frac{1}{x^2}dx$converges and $\frac{1}{x^2+1}<\frac{1}{x^2}$in the interval $[1,\infty )$, the improper integral ${\int }_1^{\infty }\frac{1}{x^2+1}dx$must converge in the interval $[1,\infty )$.