Evaluate the integrals as follows:

Substitute \(\displaystyle{u}={\ln{{x}}},{d}{u}={\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.},{v}'={1},{v}={x}\ \in\ \int{u}{d}{v}={u}{v}-\int{v}{d}{u}\).

\(\displaystyle{\int_{{{0}}}^{{{1}}}}-{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}=-{\int_{{{0}}}^{{{1}}}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}\)

\(\displaystyle=-{{\left[{x}{\ln{{\left({x}\right)}}}-\int{1}{\left.{d}{x}\right.}\right]}_{{{0}}}^{{{1}}}}\)

\(\displaystyle=-{{\left[{x}{\ln{{\left({x}\right)}}}-{x}\right]}_{{{0}}}^{{{1}}}}\)

=1

Substitute \(\displaystyle{u}={\ln{{x}}},{d}{u}={\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.},{v}'={1},{v}={x}\ \in\ \int{u}{d}{v}={u}{v}-\int{v}{d}{u}\).

\(\displaystyle{\int_{{{0}}}^{{{1}}}}-{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}=-{\int_{{{0}}}^{{{1}}}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}\)

\(\displaystyle=-{{\left[{x}{\ln{{\left({x}\right)}}}-\int{1}{\left.{d}{x}\right.}\right]}_{{{0}}}^{{{1}}}}\)

\(\displaystyle=-{{\left[{x}{\ln{{\left({x}\right)}}}-{x}\right]}_{{{0}}}^{{{1}}}}\)

=1