[tex]\large{\displaystyle \red{ \tt\int_{0}^{1} \frac{x \ln^{2}(x) }{1 + {x}^{2} } dx }}[/tex]
︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎​

Respuesta :

Step-by-step explanation:

[tex]{\displaystyle \red{ \tt = \frac{1}{2} \int_{0}^{1} \frac{x {}^{ \frac{1}{2} } \ln^{2}(x {}^{ \frac{1}{2} } ) }{1 + {x}^{} } \frac{dx}{ {x}^{ \frac{1}{2} } } }}[/tex]

[tex]{\displaystyle \red{ \tt = \frac{1}{8} \int_{0}^{1} \frac{\ln^{2}(x)}{ {1 + x}^{ \frac{}{} } }dx }}[/tex]

[tex]{\displaystyle \red{ \tt = \frac{1}{8} \sum_{k = 0}^{ \infty } ( - 1) {}^{k} \int_{0}^{1} {x}^{k} \ln^{2}(x)dx }}[/tex]

[tex]{\displaystyle \red{ \tt = \dfrac{1}{8} \sum_{k = 0}^{ \infty } ( - 1) {}^{k} \frac{( - 1 {)}^{2} 2!}{(k + 1 {)}^{3} }}}[/tex]

[tex]{\displaystyle \red{ \tt = \frac{1}{4} \sum_{k = 1}^{ \infty } \frac{( - 1 {)}^{k - 1}}{k{}^{3} }}}[/tex]

[tex]\tt \red{ = \dfrac{1}{4}\eta(3) }=[/tex]

[tex]\tt \red{ = \dfrac{3}{16} \zeta(3) }[/tex]