部分積分を利用すれば上手くいくかな？
\(I=\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}\log{\left(1+x^2\right)}dx\)
\(=\displaystyle \int_{0}^{1}\left\{\dfrac{1}{2}\log{\left(1+x^2 \right)}\right\}’\log{\left(1+x^2 \right)}dx\)
\(=\left[\dfrac{1}{2}\left\{\log{\left(1+x^2 \right)} \right\}^2 \right]_{0}^{1}-\displaystyle \int_{0}^{1}\dfrac{1}{2}\log{\left(1+x^2 \right)}\dfrac{2x}{1+x^2}dx\)
\(I=\dfrac{1}{2}\left(\log2 \right)^2-I\)
\(2I=\dfrac{1}{2}\left(\log2 \right)^2\)
∴ \(I=\dfrac{1}{4}\left(\log2 \right)^2\)・・・(答)
\(=\displaystyle \int_{0}^{1}\left\{\dfrac{1}{2}\log{\left(1+x^2 \right)}\right\}’\log{\left(1+x^2 \right)}dx\)
\(=\left[\dfrac{1}{2}\left\{\log{\left(1+x^2 \right)} \right\}^2 \right]_{0}^{1}-\displaystyle \int_{0}^{1}\dfrac{1}{2}\log{\left(1+x^2 \right)}\dfrac{2x}{1+x^2}dx\)
\(I=\dfrac{1}{2}\left(\log2 \right)^2-I\)
\(2I=\dfrac{1}{2}\left(\log2 \right)^2\)
∴ \(I=\dfrac{1}{4}\left(\log2 \right)^2\)・・・(答)
