Another fractional part integral

$I=\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} } \frac { x }{ 1-x } dx$

By Maclaurin series I'll write it as: $I=\sum _{ k=1 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} { x }^{ k }dx } }$

Let $\displaystyle J=\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} { x }^{ k }dx }$

Substitute: $\frac { 1 }{ x } =x$

$J=\sum _{ i=1 }^{ \infty }{ \int _{ 1 }^{ \infty }{ \frac { y-i }{ { y }^{ k+2 } } dy } }$

On solving it, we get: $J=\frac { 1 }{ k } -\frac { \zeta \left( k+1 \right) }{ k+1 }$

Now, coming back to our main sum: $I=\sum _{ k=1 }^{ \infty }{ \left( \frac { 1 }{ k } -\frac { \zeta \left( k+1 \right) }{ k+1 } \right) }$

I'll rearrange the terms as:

$I=\sum _{ k=1 }^{ \infty }{ \left( \frac { 1 }{ k } -\frac { 1 }{ k+1 } \right) } -\sum _{ k=1 }^{ \infty }{ \frac { \zeta \left( k+1 \right) -1 }{ k+1 } }$

Hence, $\boxed{I=\gamma }$

$\displaystyle\int_{0}^{1}{\left\{\frac{1}{x}\right\}}\frac{x}{1-x}\mathrm{d}x=\sum_{n=1}^{\infty}\int_{\frac{1}{n+1}}^{\frac{1}{n}}{\left(\frac{1}{x}-n\right)}\frac{x}{1-x}\mathrm{d}x=\sum_{n=1}^{\infty}\int_{\frac{1}{n+1}}^{\frac{1}{n}}{\left(n-\frac{n-1}{1-x}\right)\mathrm{d}x}$

$\displaystyle=\frac{1}{2}+\sum_{n=2}^{\infty}\left(n(\frac{1}{n}-\frac{1}{n+1})+(n-1)\big(\ln(1-\frac{1}{n})-\ln(1-\frac{1}{n+1})\big)\right)$

$\displaystyle=\frac{1}{2}+\sum_{n=2}^{\infty}\left(\frac{1}{n+1}+(n-1)\big(\ln(\frac{n-1}{n})-\ln(\frac{n}{n+1})\big)\right)$

$\displaystyle=\frac{1}{2}+\sum_{n=2}^{\infty}\left(\frac{1}{n+1}+(n-1)\big(\ln(n-1)-2\ln(n)+\ln(n+1)\big)\right)$

$\displaystyle=\frac{1}{2}+\lim\limits_{N \to \infty}\sum_{n=2}^{N}\left(\frac{1}{n+1}+(n-1)\big(\ln(n-1)-2\ln(n)+\ln(n+1)\big)\right)$

$\displaystyle=\lim\limits_{N \to \infty}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,+\,\,\sum_{n=2}^{N}(n-1)\ln(n-1)-2\sum_{n=2}^N(n-1)\ln(n)+\sum_{n=2}^N(n-1)\ln(n+1)$

$\displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,+\,\,\sum_{n=1}^{N-1}n\ln(n)-2\sum_{n=2}^N(n-1)\ln(n)+\sum_{n=3}^{N+1}(n-2)\ln(n)$

$\displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,+\,\,\sum_{n=3}^{N-1}n\ln(n)-2\sum_{n=3}^{N-1}(n-1)\ln(n)+\sum_{n=3}^{N-1}(n-2)\ln(n)-N\ln(N)+(N-1)\ln(N+1)$

$\displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,-\,N\ln(N)+(N-1)\ln(N+1)$

$\displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=2}^{N+1}\left(\frac{1}{n}\right)\,\,+\,\,\ln\left(\frac{(N+1)^{N-1}}{N^N}\right)$

$\displaystyle=\lim\limits_{{N \to \infty}}-1+\sum_{n=1}^{N+1}\left(\frac{1}{n}\right)\,\,+\,\,\ln\left(\frac{(N+1)^{N}}{N^N}\right)-\ln(N+1)$

$\displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N+1}\left(\frac{1}{n}\right)\,-\,\ln(N+1)\,\,+\,\,\lim\limits_{{N \to \infty}}\ln\left(\big(1+\frac{1}{N}\big)^N\right)-1$

$\displaystyle=\gamma\,\,+\,\,\ln\left(e\right)-1\,\,=\,\,\boxed{\gamma}$

See my solution...below...