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$I=\displaystyle\int x\ln(2+x^2) \, dx\\ u=\ln(2+x^2) \Rightarrow du=\dfrac{2x}{x^2+2}\\ dv=xdx \Rightarrow v=\dfrac{x^2+2}{2}\\ I=\ln(2+x^2)\dfrac{x^2+2}{2}-\displaystyle\int \dfrac{x^2+2}{2}.\dfrac{2x}{x^2+2} \, dx\\ =\ln(2+x^2)\dfrac{x^2+2}{2}-\displaystyle\int x \, dx\\ =\ln(2+x^2)\dfrac{x^2+2}{2}-\dfrac{x^2}{2}+C$
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