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Ответ:

Метод интегрирования по частям :   [tex]\bf \displaystyle \int u\cdot dv=u\cdot v-\int v\cdot du[/tex]   .

[tex]\bf \displaystyle \int ln(x^2+2)\, dx=\Big[\ u=ln(x^2+2)\ ,\ du=\frac{2x\, dx}{x^2+2}\ ,\ dv=dx\ ,\ v=x\ \Big]=\\\\\\=x\cdot ln(x^2+2)-2\int \frac{x^2\, dx}{x^2+2}=x\cdot ln(x^2+2)-2\int \Big(1-\frac{2}{x^2+2}\Big)\, dx=\\\\\\=x\cdot ln(x^2+2)-2\cdot \Big(x-\frac{2}{\sqrt2}\, arctg\, \frac{x}{\sqrt2}\Big)+C=\\\\\\=x\cdot ln(x^2+2)-2x+2\sqrt2\, arctg\, \frac{x}{\sqrt2}+C[/tex]