A cura di: Gianluca

$int_1^ex(logx)dx=$

Integrando per parti l’integrale indefinito

$x^2/2logx-int(x^2/2)(1/x)dx=$

=$x^2/2logx-int(x/2)dx$

quindi

$[(x^2/2)logx-x^2/4]_1^e=$

=$e^2/2(loge)-e^2/4-1/2(log1)+1/4=$

=$e^2/2-e^2/4+1/4=$

=$(2e^2-e^2+1)/4=$

=$(e^2+1)/4$

di Anoè Gianluca – www.chenesoio.com