Evaluer-1-logx-1-x-dx-

Question Number 179494 by a.lgnaoui last updated on 29/Oct/22

E v a l u e r

\int \frac{1 - logx}{1 + x} dx

Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22

\int \frac{1 - \log x}{1 + x} d x

= \log (1 + x) - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int x^{n} \log x d x

= \log (1 + x) - \sum_{n ⩾ 0} {(- 1)}^{n} (\frac{x^{n + 1} \log x}{n + 1} - \frac{1}{n + 1} \int x^{n} d x)

= \log (1 + x) - \sum_{n ⩾ 0} {(- 1)}^{n} (\frac{x^{n + 1} \log x}{n + 1} - \frac{x^{n + 1}}{{(n + 1)}^{2}})

Commented by a.lgnaoui last updated on 30/Oct/22

g o o d t h a n k y o u