f(x) = x⁴ + x + 1
f(x) - x⁴ -1 = x⁴ + x + 1 - x⁴ -1 = x
Integrala devine :
[tex]I=\int^e _1 xlnx\ dx[/tex]
Folosim inegrarea prin părți:
[tex]\it \int fg' = fg - \int f'g
\\\;\\
f=lnx \Rightarrow f'=\dfrac{1}{x}
\\\;\\
g'=x \Rightarrow g=\dfrac{x^2}{2}[/tex]
[tex]\it I = \dfrac{x^2}{2}lnx \Big{|}^e _1 - \int^e _1 \dfrac{1}{x}\cdot \dfrac{x^2}{2} \ dx = \dfrac{x^2lnx}{2} \Big{|}^e _1 - \int^e _1 \dfrac{x}{2}\ dx =
\\\;\\ \\\;\\
= \dfrac{x^2lnx}{2} \Big{|}^e _1 - \dfrac{x^2}{4}\Big{|}^e _1 = \dfrac{e^2\cdot1}{2} -\dfrac{1\cdot 0}{2} -\dfrac{e^2}{4} +\dfrac{1}{4} = \dfrac{e^2}{2} -\dfrac{e^4}{4} +\dfrac{1}{4} =
[/tex]
[tex]\it = \dfrac{e^2}{4} +\dfrac{1}{4} = \dfrac{e^2+1}{4}.
[/tex]
