For each integral, explain why is improper and evaluate it$.$ If needed, state "divergent."

$[-1,1]$ ase $1$

${\displaystyle \underset{-1}{\overset{0}{}}}\frac{1}{x}dx6\frac{\underset{?}{lim}}{0}{\displaystyle \underset{-1}{\overset{6}{}}}\frac{1}{x}dx\underset{60^{-}}{lim}ln|x{|}^{-1}$

$-10$

$=\underset{b{0}^{-}}{lim}(ln|b|-ln|-1|)$

${\displaystyle \underset{2}{\overset{4}{}}}\frac{1}{(x-3{)}^2}dx,cose(3)$

DNe

improfur ${\displaystyle \underset{2}{\overset{4}{}}}\frac{1}{(x-3{)}^2}dn{t}_2^3\frac{1}{(x-3{)}^2}dx+{\displaystyle \underset{3}{\overset{4}{}}}\frac{1}{(x-3{)}^2}dx$

$\{\begin{array}{c}\text{:}{\displaystyle \underset{1}{\overset{}{}}}x{e}^{-x}dx\end{array}$