Question

Ma ajutati si pe mine va rog cu o rezolvare mai detaliata la "b"? Gresesc ceva la semne:in loc de e²+1 imi iese "-1"​

Answer 1

$\displaystyle \int_{1}^e x\ln x\, dx = \int_{1}^e \Big(\dfrac{x^2}{2}\Big)'\, \ln x\, dx = \\ \\ = \dfrac{x^2\ln x}{2}\Bigg|_{1}^e - \int_{1}^e \dfrac{x^2}{2}\cdot (\ln x)'\, dx = \\ \\ = \dfrac{e^2\cdot 1}{2}-0-\int_{1}^e \dfrac{x^2}{2}\cdot \dfrac{1}{x}\, dx = \\ \\ = \dfrac{e^2}{2}- \int \dfrac{x}{2}\, dx = \\ \\ = \dfrac{e^2}{2} - \dfrac{x^2}{4}\Bigg|_{1}^e = \\ \\ = \dfrac{e^2}{2} - \dfrac{e^2}{4}+\dfrac{1}{4} = \\ \\ =\dfrac{2e^2-e^2+1}{4} = \\ \\ = \boxed{\dfrac{e^2+1}{4}}$