I'm interested in integrals of the form $$\mathfrak{I}_k=\int\left(\frac1{x^k}+\frac1{x^{k-1}(x+a)}+\cdots+\frac1{x(x+a)^{k-1}}+\frac1{(x+a)^k}\right)\,dx$$ where $a,k\in\mathbb{N}$.
We have \begin{align} \mathfrak{I}_1&=\int\left(\frac1x+\frac1{x+a}\right)\,dx=\ln(x(x+a))+C \\ \mathfrak{I}_2&=\int\left(\frac1{x^2}+\frac1{x(x+a)}+\frac1{(x+a)^2}\right)\,dx=-\left(\frac1x+\frac1{x+a}\right)\ln\left(\frac{x}{x+a}\right)+C\\ \mathfrak{I}_3&=\int\left(\frac1{x^3}+\frac1{x^2(x+a)}+\frac1{x(x+a)^2}+\frac1{(x+a)^3}\right)\,dx=-\frac12\left(\frac1x+\frac1{x+a}\right)^2+C. \end{align} But unfortunately $\mathfrak{I}_4\neq-\dfrac13\left(\dfrac1x+\dfrac1{x+a}\right)^3+C$ breaks the pattern according to Wolfram.
So the question is:
Is there a closed form to the integral $$\mathfrak{I}_k=\int\sum_{n=0}^k\frac1{x^{k-n}(x+a)^n}\,dx \text{ for } a,k\in\mathbb{N}?$$
The $a$-parameter is irrelevant since it can be removed through the substitution $x=az$, so let us focus on
$$I_k = \int \sum_{j=0}^k \frac{1}{x^j(x+1)^{k-j}}\,dx = \int\frac{x+1}{x^k}-\frac{x}{(x+1)^k}\,dx $$ which is an elementary integral, equal to $$C+\frac{x^{2-k}-(x+1)^{2-k}}{2-k}+\frac{x^{1-k}-(x+1)^{1-k}}{1-k} $$ for any $k\geq 3$. Mystery solved.