Let $f$ be the function defined by
$$f(x) = \sum_{n=1}^{+\infty} \left(\frac{1}{n+x} - \frac{1}{n}\right)$$
By noting that, for all $t \in [0,1[$, $\frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}$, show that
$$\forall x \in {]-1,+\infty[}, \quad f(x) = \int_{0}^{1} \frac{t^{x} - 1}{1 - t} \mathrm{~d}t$$