Is there a function f: R->R is integrable but f^2 is not? why?
Theorem: If $\displaystyle f$ is an integrable function in $\displaystyle [a,b]$ then $\displaystyle f^2$ is an integrable function in $\displaystyle [a,b]$.
P.S. In response to the to reply #3.
It depends on how one defines integral.
Most often Riemann integrals require bounded functions.
That example is not a bounced function.
The term is improper integral.
Consider $\displaystyle f(x) =\begin{cases} \frac 1{\sqrt{|x|}}&\mbox{if } -1\leq x<0 \mbox{ or }0<x\leq 1\\0&\mbox{ if } |x|>1 \end{cases}$ and choose any value you want for $\displaystyle x=0$.
The function $\displaystyle f$ is integrable but not its square. As you noticed, a primitive of $\displaystyle \frac 1x$ is $\displaystyle \ln x$, but if you compute $\displaystyle \int_{\varepsilon}^1\frac 1xdx$, you will find $\displaystyle -\ln \varepsilon$, and it has an infinite limit as $\displaystyle \varepsilon\to 0$.
Read my P.S.
It depends on what we mean by integrable.
The function $\displaystyle \frac{1}{\sqrt{|x|}}$ is not Riemann integrable on $\displaystyle [-1,0]$.
However as an improper integral $\displaystyle \int_{ - 1}^0 {\frac{{dx}}{{\sqrt {\left| x \right|} }}}=2 $.
But $\displaystyle \int_{ - 1}^0 {\frac{{dx}}{{ {\left| x \right|} }}}$ does not exist even as an improper integral.