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find-0-lnx-x-2-x-1-dx-




Question Number 118955 by Bird last updated on 21/Oct/20
find ∫_0 ^∞   ((lnx)/(x^2 +x+1))dx
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$
Answered by mnjuly1970 last updated on 21/Oct/20
  I=∫_0 ^( ∞) ((ln(x))/(1+x+x^2 ))dx=^(x=(1/t)) −∫_0 ^( ∞) ((ln(t))/(t^2 +t+1))  2I=0⇒I=0  ✓
$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{dx}\overset{{x}=\frac{\mathrm{1}}{{t}}} {=}−\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +{t}+\mathrm{1}} \\ $$$$\mathrm{2}{I}=\mathrm{0}\Rightarrow{I}=\mathrm{0}\:\:\checkmark \\ $$$$ \\ $$

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