Ma poate ajuta cineva cu această integrală?

[tex]\texttt{Se desparte in doua :}\\\displaystyle\int_1^e\left(x+\dfrac{1}{x}\right)\ln xdx=\int_1^ex\ln xdx+\int_1^e\dfrac{1}{x}\cdot\ln xdx\\\texttt{Le luam separat:}\\\int_1^ex\ln xdx=\dfrac{x^2}{2}\cdot\ln x|_1^e-\int_1^e\dfrac{x}{2}dx=\dfrac{e^2}{2}-\dfrac{x^2}{4}|_1^e=\dfrac{e^2}{2}-\dfrac{e^2}{4}+\dfrac{1}{4}=\dfrac{e^2}{4}+\dfrac{1}{4}\\\texttt{Pentru a doua integrala facem o schimbare de variabila:}\\\ln x=t\\\dfrac{dx}{x}=dt[/tex]

[tex]\displaystyle\int_1^e\dfrac{1}{x}\cdot\ln xdx=\int_0^1tdt=\dfrac{t^2}{2}|_0^1=\dfrac{1}{2}\\\texttt{Asadar:}\\\int_1^e\left(x+\dfrac{1}{x}\right)\cdot\ln xdx=\dfrac{e^2}{4}+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{e^2}{4}+\dfrac{3}{4}=\dfrac{e^2+3}{4}[/tex]

Rayzen Rayzen · Answer 2 · 2019-06-06T17:25:37+03:00

[tex]I = \displaystyle \int_{1}^e \Big(x+\frac{1}{x}\Big)\ln x\, dx= \int_{1}^e\dfrac{1}{x}\Big(x^2+1\Big)\ln x\, dx \\ \\ \ln x = t \Rightarrow \dfrac{1}{x}\, dx = dt \\ x = e^t \\ \\ x = 1 \Rightarrow t = 0,\quad x = e \Rightarrow t = 1 \\ \\ I = \int_{0}^1(e^{2t}+1)t\, dt = \dfrac{1}{2}\int_{0}^1(e^{2t})'t\, dt+\int_{0}^1 t\, dt = \\ \\\\ = \dfrac{e^{2t}t}{2}\Big|_0^1-\dfrac{1}{2}\int e^{2t}\, dt+\dfrac{t^2}{2}\Big|_0^1 =[/tex]

[tex]= \dfrac{e^2}{2}-\dfrac{1}{2\cdot 2}\cdot e^{2t}\Big|_{0}^1+\dfrac{1}{2} = \dfrac{e^2}{2}-\dfrac{e^2}{4}+\dfrac{1}{4}+\dfrac{1}{2} = \boxed{\dfrac{e^2+3}{4}}[/tex]