calculate-1-2-logx-1-x-dx-

Question Number 148564 by mathmax by abdo last updated on 29/Jul/21

calculate \int_{1}^{2} \frac{logx}{1 + x} dx

Answered by Kamel last updated on 29/Jul/21

Ω = \int_{1}^{2} \frac{L n (x)}{1 + x} d x = - \int_{\frac{1}{2}}^{1} \frac{L n (x)}{x (1 + x)} d x

= \frac{1}{2} {L n}^{2} (2) + L n (2) L n (\frac{3}{2}) - \int_{\frac{1}{2}}^{1} \frac{L n (1 + x)}{x} d x

= L n (2) L n (3) - \frac{1}{2} {L n}^{2} (2) + {L i}_{2} (- 1) - {L i}_{2} (- \frac{1}{2})

∴ \int_{1}^{2} \frac{L n (x)}{1 + x} d x = L n (2) L n (3) - \frac{1}{2} {L n}^{2} (2) - \frac{π^{2}}{12} - {L i}_{2} (- \frac{1}{2})

Facebook Tweet Pin