Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises.

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

Short answer

The improper integral ${\int }_1^{\infty }\frac{x+1}{x^2}dx$diverges.

Step-by-step solution

Step 1. Given information. 

The given integral is the following.

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

Step 2. Value of integral. 

Consider function$f\left(x\right)=\frac{x+1}{x^2}$

localid="1648879879677" $$\begin{gathered} f\left(x\right)=\frac{x+1}{x^2} \\ =\frac{x}{x^2}+\frac{1}{x^2} \\ =\frac{1}{x}+\frac{1}{x^2} \\ \frac{1}{x}+\frac{1}{x^2}>\frac{1}{x^2} \end{gathered}$$

In the interval $[1,\infty )$we have localid="1648879786507" $\frac{x+1}{x^2}>\frac{1}{x^2}.$

As we know that localid="1648884171467" ${\int }_1^{\infty }\frac{1}{x^2}dx$is converse and integral of functions less than $\frac{1}{x^2}$will be converse.

But $\frac{x+1}{x^2}\nless \frac{1}{x^2}$so it cannot be converse.

so$\frac{1}{x}+\frac{1}{x^2}$is diverse.