Question Number 192712 by cortano12 last updated on 25/May/23
$$\:\:\:\:\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{dx}\:=? \\ $$
Answered by horsebrand11 last updated on 25/May/23
![I=2∫_0 ^(1/2) (√(x^2 +1+(√((x^2 +x+1)(x^2 −x+1))))) dx = (√2) ∫_0 ^(1/2) (√(((√(x^2 +x+1)))+(√(x^2 −x+1)) )^2 )) dx = (√2) ∫_0 ^(1/2) ((√(x^2 +x+1)) +(√(x^2 −x+1)) )dx =(1/( (√2) ))(∫_0 ^(1/2) (√((2x+1)^2 +3)) dx +∫_(−1/2) ^0 (√((2x+1)^2 +3)) dx) = (1/( (√2))) ∫_(−1/2) ^(1/2) (√((2x+1)^2 +3)) dx let 2x+1=(√3) tan α I= (3/(2(√2))) ∫_0 ^(arctan (2/(√(3)))) sec^3 α dα = (3/(4(√2))) (sec α tan α +ln (sec α+tan α) ]_0 ^(arctan ((2/( (√3))))) =2(√7) +3ln (((2+(√7))/( (√3))))](Q192714.png)
$${I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}}\:{dx} \\ $$$$\:=\:\sqrt{\mathrm{2}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\sqrt{\left.\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\right)^{\mathrm{2}} }\:{dx} \\ $$$$\:=\:\sqrt{\mathrm{2}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\right){dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\left(\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}}\:{dx}\:+\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{0}} {\int}}\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}}\:{dx}\right) \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}}\:{dx}\: \\ $$$$\:{let}\:\mathrm{2}{x}+\mathrm{1}=\sqrt{\mathrm{3}}\:\mathrm{tan}\:\alpha\: \\ $$$${I}=\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{\mathrm{0}} {\overset{\mathrm{arctan}\:\left(\mathrm{2}/\sqrt{\left.\mathrm{3}\right)}\right.} {\int}}\mathrm{sec}\:^{\mathrm{3}} \alpha\:{d}\alpha\: \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\left(\mathrm{sec}\:\alpha\:\mathrm{tan}\:\alpha\:+\mathrm{ln}\:\left(\mathrm{sec}\:\alpha+\mathrm{tan}\:\alpha\right)\:\right]_{\mathrm{0}} ^{\mathrm{arctan}\:\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)} \\ $$$$=\mathrm{2}\sqrt{\mathrm{7}}\:+\mathrm{3ln}\:\left(\frac{\mathrm{2}+\sqrt{\mathrm{7}}}{\:\sqrt{\mathrm{3}}}\right)\: \\ $$