logarithmic Integral

\hspace{-16}\mathbf{\int_{0}^{1}\frac{\ln(1+x)}{x}dx=\int_{0}^{1}\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+.......+\infty}{x}dx}\\\\\\ $Using Series Formula of $\mathbf{\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+.......+\infty}$\\\\\\ So $\mathbf{\int_{0}^{1}\frac{\ln(1+x)}{x}dx=\int_{0}^{1}\left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+.....+\infty\right)dx}$\\\\\\ $\mathbf{\int_{0}^{1}\frac{\ln(1+x)}{x}dx=1-\frac{1}{2^2}+\frac{1}{3^2}-....+\infty=\frac{\pi^2}{12}}$\\\\\\ Bcz We Know That $\mathbf{1+\frac{1}{2^2}+\frac{1}{3^2}+....+\infty=\frac{\pi^2}{6}}$\\\\\\ $\mathbf{\left(1+\frac{1}{3^2}+\frac{1}{5^2}+..+\infty\right)+\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+..+\infty\right)=\frac{\pi^2}{6}}$\\\\\\ So $\mathbf{\left(1+\frac{1}{3^2}+\frac{1}{5^2}+..+\infty\right)=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}}$\\\\\\ So $\mathbf{1-\frac{1}{2^2}+\frac{1}{3^2}-....+\infty}$\\\\\\ $\mathbf{=}$
\hspace{-16}\mathbf{\left(1+\frac{1}{3^2}+\frac{1}{5^2}+..+\infty\right)-\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+..+\infty\right)}$\\\\\\ $\mathbf{=\frac{\pi^2}{8}-\frac{1}{2^2}.\frac{\pi^2}{6}=\frac{\pi^2}{12}}$\\\\\\ So $\mathbf{I_{1}=\int_{0}^{1}\frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}}$\\\\\\ $\boxed{\mathbf{\frac{1}{I_{1}}=\frac{12}{\pi^2}}}$\\\\\\ Similarly\\\\\\ $\mathbf{\int_{0}^{1}\frac{\ln(1+x)}{x}dx=\int_{0}^{1}\frac{-x-\frac{x^2}{2}-\frac{x^3}{3}-.......-\infty}{x}dx}$\\\\\\ So $\mathbf{\int_{0}^{1}\frac{\ln(1-x)}{x}dx=\int_{0}^{1}\left(-1-\frac{x}{2}-\frac{x^2}{3}-\frac{x^3}{4}-.....-\infty\right)dx}$\\\\\\ $\mathbf{\int_{0}^{1}\frac{\ln(1-x)}{x}dx=-1-\frac{1}{2^2}-\frac{1}{3^2}-....-\infty=-\frac{\pi^2}{6}}$\\\\\\ So $\mathbf{I_{2}=\int_{0}^{1}\frac{\ln(1-x)}{x}dx=-\frac{\pi^2}{6}}$\\\\\\ $\boxed{\mathbf{\frac{1}{I_{2}}=-\frac{6}{\pi^2}}}$\\\\\\ So $\mathbf{\left(\frac{1}{I_{1}}+\frac{1}{I_{1}}\right).\pi^2=\left(\frac{12}{\pi^2}-\frac{6}{\pi^2}\right).\pi^2=6}$