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Question Number 80175 by Rio Michael last updated on 31/Jan/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:=\:? \\$$

Commented bymathmax by abdo last updated on 31/Jan/20

$${let}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\$$ $${cha}\mathrm{7}{gement}\:\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}\theta\:{give}\:\:\theta\:={arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right) \\$$ $${I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta\:=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{d}\theta \\$$ $$=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{d}\theta}{{cos}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={u}} \:\:\:\:\int_{\mathrm{2}−\sqrt{\mathrm{3}}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}×\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\$$ $$=\int_{\mathrm{2}−\sqrt{\mathrm{3}}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \left(\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du}\:=\left[{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\right]_{\mathrm{2}−\sqrt{\mathrm{3}}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \\$$ $$={ln}\mid\frac{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\mid−{ln}\mid\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{−\mathrm{1}+\sqrt{\mathrm{3}}}\mid={ln}\mid\sqrt{\mathrm{3}}+\mathrm{1}\mid−{ln}\mid\sqrt{\mathrm{3}}−\mathrm{1}\mid−{ln}\mid\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\sqrt{\mathrm{3}}−\mathrm{1}}\mid \\$$ $$={ln}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)−{ln}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)−{ln}\left(\sqrt{\mathrm{3}}\right) \\$$

Answered by ~blr237~ last updated on 31/Jan/20

$${Let}\:{named}\:{it}\:{A} \\$$ $${using}\:\:\mathrm{1}+{x}+{x}^{\mathrm{2}} =\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\left[\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}\right] \\$$ $$\\$$ $${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\:\frac{\mathrm{1}}{\sqrt{\left[\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}\right.}}\:{dx} \\$$ $$\:\:=\left[{argsh}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\:{argsh}\left(\sqrt{\mathrm{3}}\:\right)−{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right) \\$$ $${using}\:\:{argshx}={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)\: \\$$ $${A}={ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)−{ln}\left(\sqrt{\mathrm{3}}\:\right) \\$$ $$\: \\$$