Evaluation of $$\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$$
Let $$I = \int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx = \int^{1}_{0}\frac{(1-x)+\ln x}{(1-x)\ln x}dx$$
Now How can i solve after that , Help required, Thanks
Evaluation of $$\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$$
Let $$I = \int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx = \int^{1}_{0}\frac{(1-x)+\ln x}{(1-x)\ln x}dx$$
Now How can i solve after that , Help required, Thanks
It is actually know that this integral is equal to the Euler Mascheroni constant.
$$\gamma=\int _0^1\left ({1\over1-x}+{1\over\ln x}\right)dx\approx0.57721566490153286060651209008240243104215933593992$$