Sub integrală avem :
[tex]\it (f(x)-x^4-1)lnx= (x^4+x+1-x^4-1)lnx =xlnx
\\ \\
I = \int^e_1 xlnx dx[/tex]
Folosim integrarea prin părți:
[tex]\it \int fg'=fg-\int f'g
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f= lnx \Rightarrow f'=\dfrac{1}{x}
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g'=x \Rightarrow g=\int x dx \Rightarrow g = \dfrac{x^2}{2}[/tex]
[tex]\it I = \dfrac{x^2}{2}lnx \Big{|}^e_1 \ - \ \int^e _1 \dfrac{1}{x} \dfrac{x^2}{2} dx = \dfrac{x^2}{2}lnx \Big{|}^e_1 \ - \ \int^e _1 \ \dfrac{x}{2} dx =
\\ \\ \\
= \dfrac{x^2}{2}lnx \Big{|}^e_1 \ - \ \dfrac{x^2}{4}\Big{|}^e_1 = \dfrac{e^2}{2}-0-\left(\dfrac{e^2}{4} -\dfrac{1}{4}\right) = \dfrac{^{2)}e^2}{\ 2}-\dfrac{e^2}{4} +\dfrac{1}{4} =
\\ \\ \\ = \dfrac{2e^2-e^2+1}{4} = \dfrac{e^2+1}{4}[/tex]
