A cura di: Francesco Speciale
Svolgimento:
$int_(1)^(3)(1/(x(1+x)))dx=int_(1)^(3)((1+x-x)/(x(1+x)))dx=$
$=int_(1)^(3)(1/x)dx-int_(1)^(3)(1/(1+x))dx=$
$[log(|x|)-log(|1+x|)]_(1)^(3)=log(1)-log(2)-(log(3)-log(4))=log(4)-log(3)+log(2)$.
- Integrali