I have been asked to explain whether the following is Riemann integrable:
On
The function $f$ is Riemann-integrable, but your justification doesn't work. It is not true that every bounded function is Riemann-integrable; take $\chi_{\mathbb{Q}\cap[0,1]}\colon[0,1]\longrightarrow\mathbb R$, for instance.
The function $f$ is Riemann-integrable because it is bounded and it is discontinuous only at a single point (which is $\frac14$).
On
In general you can note down the following two theorems which don't require measure theory for their proof.
Theorem 1: If $f$ is continuous on $[a, b] $ then $f$ is Riemann integrable on $[a, b] $.
Theorem 2: If $f$ is monotone on $[a, b] $ then $f$ is Riemann integrable on $[a, b] $.
The first theorem can be generalized further (again without recourse to measure theory) to the following:
Theorem 3: If $f$ is bounded on $[a, b] $ and the set $D$ of discontinuities of $f$ on $[a, b] $ has only a finite number of limit points then $f$ is Riemann integrable on $[a, b] $.
An immediate consequence of the above theorem is that $f$ is Riemann integrable integrable if $f$ is bounded and the set $D$ of its discontinuities is finite. And that's what we need here. The given function is bounded and discontinuous at just a single point and therefore Riemann integrable.
Try writing $$ \int_{-1}^1f(x)\;\mathrm dx=\int_{-1}^{1/4}x\;\mathrm dx+\int_{1/4}^1\frac{1}{x}\;\mathrm dx $$ Both pieces we know are integrable, indeed they are bounded and continuous.
It would probably be good to note that the integral doesn't care about discontinuities at a (or countably many) point(s).
Edit: A little more justification for why we don't need to care about the contribution at $\frac{1}{4}$, note that $$ \left|\int_{1/4-\epsilon/8}^{1/4+\epsilon/8}f(x)\;\mathrm dx\right |\leq\epsilon/4\sup_{-1\leq x\leq 1}|f(x)|=\epsilon\to 0 $$ as we shrink $\epsilon \to 0$.