Evaluation of $\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$

Question

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

Is there any reason to expect a reasonably nice closed form? — Daniel Fischer, Nov 25 '16 at 14:22

score 2 · Answer 1 · answered Nov 25 '16 at 17:08

2

It is actually know that this integral is equal to the Euler Mascheroni constant.

$$\gamma=\int _0^1\left ({1\over1-x}+{1\over\ln x}\right)dx\approx0.57721566490153286060651209008240243104215933593992$$

answered Nov 25 '16 at 17:08

1 Answers1