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
- Integrali