The integral $\int_0^1 \frac{(x+1)^n-1}{x} dx$

I know that the integral $\int_0^1 \frac{(x+1)^n-1}{x} dx,$ for $n \in \mathbb{Z}^+$, can be evaluated by expanding the numerator with the binomial theorem and integrating term by term. You get the nice expression $$\int_0^1 \frac{(x+1)^n-1}{x} dx = \sum_{k=1}^n \binom{n}{k} \frac{1}{k}.$$ My question is this:

Is there some other "nice" expression for $$\int_0^1 \frac{(x+1)^n-1}{x} dx,$$ when $n$ is a positive integer?

Mathematica and Wolfram Alpha both tell me that $$\int_0^1 \frac{(x+1)^n-1}{x} dx = n \text{HypergeometricPFQ}[\{1, 1, 1 - n\}, \{2, 2\}, -1],$$ but the hypergeometric function is just another way of writing $\sum \limits_{k=1}^n \binom{n}{k} \frac{1}{k}.$

I'm led to think that there might be such a nice expression because we do get one for the same integrand but slightly different bounds: $$\int_{-1}^0 \frac{(x+1)^n-1}{x} dx = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^{k+1}}{k} = H_n,$$ where $H_n$ is the $n$th harmonic number. On the other hand, alternating binomial sums often evaluate to simpler expressions than the corresponding sums that don't alternate, so perhaps the existence of the simpler expression here doesn't mean much for my question.

I would accept a known special function as an answer (other than the hypergeometric one I already mention), or a simple expression in terms of well-known numbers, like the harmonic numbers.

Substitute $x = u-1,$ so $u = x+1,$ I get $$ \int_1^2 \; \; \frac{u^n -1}{u-1} \; du \; = \; \int_1^2 \; \; u^{n-1} + u^{n-2} + \cdots + u^2 + u^2 + u + 1 \; \; du $$ or $$ \left. \frac{u^n}{n} + \frac{u^{n-1}}{n-1} + \frac{u^{n-2}}{n-2} + \cdots + \frac{u^3}{3} + \frac{u^2}{2} + u \; \; \right|_{u=1}^{u=2} $$ The value at $u=1$ is $H_n.$ The value at $u=2$ is, well, just one of those things.

Let $J_n$ be your integral, and define $J_0 = 0$. The generating function for this sequence is $$G(z) = \sum_{n=1}^\infty J_n z^n = \frac{\ln((1-2t)/(1-t))}{t(t-1)} $$ The generating function satisfies the differential equation $$(1-2z)(1-z)^2 \frac{dy}{dz} - (1-2z)(1-z) y = 1 $$ and the coefficients satisfy the recurrence $$ (2n+2) J_n - (3n+4) J_{n+1} + (n+2) J_{n+2} = 1 $$ A "closed-form" formula obtained by having Maple solve this recurrence is $$J_n =-i\pi -{\frac {{2}^{n} \left( {\it LerchPhi} \left( 2,1,n \right) n-1 \right) }{n}}-\Psi \left( n+1 \right) -\gamma $$