I have to determine the intervals $[a,b], a < b$, where the following function is Riemann integrable:
$$f(x)=\frac{\sin x}{x^2 + x}$$ with $x \neq 0,-1, f(0)=2, f(-1)=-4$.
If $f$ is continuous in $[a,b]$, then $f$ will be Riemann integrable in $[a,b]$.
$\lim_{x\to 0} \frac{\sin x}{x^2 + x} = 1 \neq f(0) = 2$. Then $f$ is not continuous in $x=0$.
$\lim_{x\to -1} \frac{\sin x}{x^2 + x} = \infty \neq f(-1) = -4$. Then $f$ is not continuous in $x=-1$.
Now, we can say that $f$ is continuous in $x \in \Bbb R - \{-1,0\}$. Then $f$ will be Riemann integrable in $(-\infty,-1) \cup (-1,0) \cup (0,+\infty)~$.
Am I wrong? What about the points where $f$ is not continuous?
Yes, you are wrong. The conclusion is that it is integrable in $[a,b]$ if and only if $b<-1$ or $a>-1$. The discontinuity at $0$ is irrelevant. On the other hand, if $-1\in[a,b]$, then $f|_{[a,b]}$ is unbounded and therefore not Riemann-integrable.