0-1-ln-x-x-1-dx-

Question Number 118246 by bemath last updated on 16/Oct/20

\underset{0}{\int^{1}} \frac{\ln x}{x + 1} d x = ?

Answered by Dwaipayan Shikari last updated on 16/Oct/20

\int_{0}^{1} \frac{l o g x}{x + 1} d x = {[l o g (x) l o g (x + 1)]}_{0}^{1} - \int_{0}^{1} \frac{1}{x} l o g (x + 1)

= 0 - \int_{0}^{1} {(- 1)}^{n + 1} \underset{n = 1}{\sum^{\infty}} \frac{x^{n - 1}}{n}

= - \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n^{2}}

= - \frac{π^{2}}{12}