Question

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

Answer 1

${\displaystyle \red{ \tt I = \int_{0}^{1} \frac{x \ln^{2}(x) }{1 + {x}^{2} } dx }}$

${\displaystyle \red{ \tt = \frac{1}{2} \int_{0}^{1} \frac{x {}^{ \frac{1}{2} } \ln^{2}(x {}^{ \frac{1}{2} } ) }{1 + {x}^{} } \frac{dx}{ {x}^{ \frac{1}{2} } } }}$

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

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

${\displaystyle \red{ \tt = \frac{1}{8} \sum_{k = 0}^{ \infty } ( - 1) {}^{k} \frac{( - 1 {)}^{2} 2!}{(k + 1 {)}^{3} }}}$

${\displaystyle \red{ \tt = \frac{1}{4} \sum_{k = 1}^{ \infty } \frac{( - 1 {)}^{k - 1}}{k{}^{3} }}}$

$\tt \red{ = \dfrac{1}{4}\eta(3) }$

$\tt \red{ = \dfrac{3}{16} \zeta(3) }$