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Question Number 118955 by Bird last updated on 21/Oct/20

$$find{\int }_0^{\mathrm{\infty }}\frac{lnx}{x^2+x+1}dx$$

Answered by mnjuly1970 last updated on 21/Oct/20

$$$$$$I={\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x+x^2}dx\stackrel{x=\frac{1}{t}}{=}-{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)}{t^2+t+1}$$$$2I=0\Rightarrow I=0✓$$$$$$

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