Problem source

If 
\[
\mathrm{f}(x)=nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n}\,,
\]
show that 
\[
\mathrm{f}'(x)=\frac{1-(1-x)^{n}}{x}\,.
\]
Deduce that 
\[
\mathrm{f}(x)=\int_{1-x}^{1}\frac{1-y^{n}}{1-y}\,\mathrm{d}y.
\]
Hence show that 
\[
\mathrm{f}(1)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\,.
\]

Solution source

\begin{align*}
    f(x) & =nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n} \\
    f'(x) &= n - \binom{n}{2} x + \binom{n}{3}x^2 - \cdots (-1)^{r+1} \binom{n}{r} + \cdots + (-1)^{n+1} x^{n-1} \\
    &= \frac{1-(1-x)^n}{x}
\end{align*}
Therefore, since $\displaystyle f(x) = \int_0^xf'(t)\,dt$
\begin{align*}
    f(x) &= \int_0^x \frac{1 - (1-t)^n}{t} \, dt \\
    &= \int_{1}^{1-x} \frac{1-y^n}{1-y} (-1)\, dy \tag{Let $y = 1-t, \frac{dy}{dt} = -1$} \\
    &= \boxed{\int_{1-x}^1 \frac{1-y^n}{1-y} dy} \\
    &= \int_{1-x}^1 \l 1 + y + y^2 + \cdots + y^{n-1} \r \, dy \\
    &= \left [ y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots + \frac{y^n}{n} \right]_{1-x}^1 \\
\end{align*}

So when $x = 1, 1-x = 0$ so we exactly have the sum required.