Is there a nice general formula for $\int \frac{dx}{x^n-1}$ and/or $\int \frac{dx}{\Phi_n(x)}$?

You can get a pretty simple expression using partial fractions over $\mathbb{C}$. In general, if $f(x)=(x-a_1)\dots(x-a_n)$ is a monic polynomial with distinct roots over $\mathbb{C}$, then we have the partial fractions decomposition $$\frac{1}{f(x)}=\sum_{k=1}^n\frac{1}{f'(a_k)(x-a_k)}.$$ (You can verify that this decomposition is correct by multiplying both sides by $x-a_k$ and taking the limit as $x\to a_k$: on the left side you have the limit of $\frac{x-a_k}{f(x)}$ which is exactly the reciprocal of the difference quotient for $f'(a_k)$ and so converges to $\frac{1}{f'(a_k)}$.)

Thus an antiderivative can be computed as $$\int\frac{1}{f(x)}dx=\sum_{k=1}^n\frac{\log(x-a_k)}{f'(a_k)}+C.$$ You can apply this to either $f(x)=x^n-1$ or $f(x)=\Phi_n(x)$. So, when you write everything over $\mathbb{C}$ and do not attempt to combine terms to get something that is obviously real-valued, the coefficients just come from the reciprocal of the derivative of the polynomial evaluated at each root. In the case of $x^n-1$, at least, you can write this quite explicitly since the derivative is easy to evaluate: $$\int\frac{1}{x^n-1}dx=\frac{1}{n}\sum_{k=1}^n\zeta^k\log(x-\zeta^k)+C$$ where $\zeta$ is a primitive $n$th root of unity (here the $f'(\zeta^k)$ in the denominator became $n(\zeta^k)^{n-1}=n(\zeta^k)^{-1}$).

In terms of a general formula $$ \displaystyle \int \frac{dx}{x^n-1} = -x\;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x^n\right) $$ which contains a hypergeometric function. I have played around with the more general cyclotomic case but did not get anywhere in terms of hypergeometric functions.

We have in general from a geometric series $$ \frac{1}{x^n-1} = -1 - x^n -x^{2n} -x^{3n} -\cdots = -\sum_{k=1}^\infty x^{n k} $$ if we integrate term by term then we have $$ \int \frac{1}{x^n-1}\;dx = -x - \frac{x^{n+1}}{n+1} -\frac{x^{2n+1}}{2n+1} -\frac{x^{3n+1}}{3n+1} -\cdots = - \sum_{k=1}^\infty \frac{x^{kn+1}}{kn+1} $$ $$ \int \frac{1}{x^n-1}\;dx = - x\sum_{k=1}^\infty \frac{x^{kn}}{kn+1} $$ If we consider the definition of the hypergeometric function $$ \;_2F_1(a,b;c,x) = \sum_{k=1}^\infty \frac{(a)_k (b)_k x^k}{(c)_k k!} $$ with $(\cdot)_k$ the Pochhammer symbol. For this example we have $$ \;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x\right) = \sum_{k=1}^\infty \frac{\left(1\right)_k \left(\frac{1}{n}\right)_k x^k}{\left(1+\frac{1}{n}\right)_k k!} $$ we have that $(1)_k=k!$ so $$ \;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x\right) = \sum_{k=1}^\infty \frac{ \left(\frac{1}{n}\right)_k x^k}{\left(1+\frac{1}{n}\right)_k} =\sum_{k=1}^\infty \frac{x^{kn}}{kn+1} $$