The integrand has a vertical asymptote at $x\=0$. The integral diverges since

${\int }_0^1{x}^{-1} \mathit{\ⅆ}x$ = $\underset{s\to 0\+}{lim}{\int }_s^1{x}^{-1} \mathit{\ⅆ}x$ = $\underset{s\to 0\+}{lim}\left({\left[\ln \left(x\right)\right]}_s^1\right)$ = $\underset{c\to 0\+}{lim}\left(0-\ln \left(s\right)\right)\=-\left(-\infty \right)\=\infty$ 

Divergence of the integral is established if, to the right of the asymptote, the integral is divergent.

It is completely incorrect to integrate across the vertical asymptote. Hence, the following "solution" is totally wrong.

${\int }_{-1}^1{x}^{-1} \mathit{\ⅆ}x$ = ${\left[\ln \left(\left|x\right|\right)\right]}_{-1}^1$  = $\ln \left(1\right)-\ln \left(\left|-1\right|\right)\=\ln \left(1\right)-\ln \left(1\right)\=0-0\=0$ 

The only way the value of zero can be attached to this integral is via the Cauchy Principal Value, computed as follows.

The positive quantity $s$ is treated as a small variable on either side of the asymptote, the integrals on either side of the asymptote stopping at this small distance from the discontinuity. The value of the sum of the resulting integrals is then obtained in the limit as $s$ is squeezed to zero simultaneously (i.e., symmetrically) on either side of the asymptote.

The CPV can only be obtained by typing the int command with the CPV option, a shown.