\[{\text{Вычислите }}\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right)}}{x}dx} .\]
\[\begin{array}{l}
\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right)}}{x}dx} = \int\limits_0^1 {\left( {1 - \frac{1}{2}x + \frac{1}{3}{x^2} - \frac{1}{4}{x^3} + ...} \right)dx} = \left. {\left( {x - \frac{1}{4}{x^2} + \frac{1}{9}{x^3} - \frac{1}{{16}}{x^4} + ...} \right)} \right|_0^1 = \hfill \\
1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{{16}} + ... = \eta \left( 2 \right) = \frac{1}{2}\zeta \left( 2 \right) = \frac{1}{2} \cdot \frac{{{\pi ^2}}}{6} = \frac{{{\pi ^2}}}{{12}}. \hfill \\
\end{array}\]
\[\frac{{{\pi ^2}}}{{12}}\]