${\int }_1^{\infty }\frac{1}{x^2} \mathit{\ⅆ}x \=$

$\underset{t\to \infty }{lim}{\int }_1^t\frac{1}{x^2} \mathit{\ⅆ}x$

$\=$

$\underset{t\to \infty }{lim}\left({\left[-\frac{1}{x}\right]}_1^t\right)$

$\=$

$\underset{t\to \infty }{lim}\left(1-\frac{1}{t}\right)$

$\=$

$1$

Apply Maple to the improper integral

${\int }_1^{\infty }\frac{1}{x^2} \mathit{\ⅆ}x$ = $1$$$

Integrate to a finite endpoint, then take the limit

${\int }_1^t\frac{1}{x^2} \mathit{\ⅆ}x$$\stackrel{\text{assuming positive}}{\to }$$\frac{t-1}{t}$$\stackrel{\text{limit}}{\to }$$1$

Alternate evaluation of ${\int }_1^t\frac{1}{x^2} \mathit{\ⅆ}x$ 

${\int }_1^t\frac{1}{x^2} \mathit{\ⅆ}x assuming t\>1$ = $\frac{t-1}{t}$$$