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Let $n$ be a positive integer. For a real number $x$, let $[x]$ denote the largest integer not exceeding $x$ and $\{x\}=$ $x-[x] .$ Then $\int_{1}^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$ is equal to
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The correct answer is:
$\frac{\mathrm{n}}{\mathrm{n}+1}$
$\int_{1}^{n+1} \frac{\{x\}^{[x]}}{[x]} d x=\int_{1}^{2} \frac{\{x\}^{[x]}}{[x]} d x+\int_{2}^{3} \frac{\{x\}^{[x]}}{[x]} d x+\ldots .+\int_{1}^{n+1} \frac{\{x\}^{[x]}}{[x]} d x$
$=\sum_{r=1}^{n} \int_{r}^{r+1} \frac{\{x\}^{[x]}}{[x]} d x$
$=\sum_{r=1}^{n} \int_{r}^{r+1} \frac{(x-r)^{r}}{r} d x$
$=\sum_{r=1}^{n}\left[\frac{(x-r)^{r+1}}{r(r+1)}\right]_{r}^{r+1}$
$=\sum_{r=1}^{n} \frac{1}{r(r+1)}=\frac{n}{n+1}$
$=\sum_{r=1}^{n} \int_{r}^{r+1} \frac{\{x\}^{[x]}}{[x]} d x$
$=\sum_{r=1}^{n} \int_{r}^{r+1} \frac{(x-r)^{r}}{r} d x$
$=\sum_{r=1}^{n}\left[\frac{(x-r)^{r+1}}{r(r+1)}\right]_{r}^{r+1}$
$=\sum_{r=1}^{n} \frac{1}{r(r+1)}=\frac{n}{n+1}$
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