Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 101271 by mathmax by abdo last updated on 01/Jul/20

find ∫   ((xdx)/((√(x^2 +x+1))+(√(x^2 −x+1))))

$$\mathrm{find}\:\int\:\:\:\frac{\mathrm{xdx}}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}} \\ $$

Commented by Dwaipayan Shikari last updated on 01/Jul/20

∫((x((√(x^2 +x+1))−(√(x^2 −x+1))))/(2x))  (1/2)∫(√(x^2 +x+1))−(1/2)∫(√(x^2 −x+1))  (1/2)∫(√((x+(1/2))^2 +(((√3)/2))^2 ))dx−(1/2)∫(√((x−(1/2))^2 +(((√3)/2))^2    )){suppose x+(1/2)=t  (1/2)∫(√(t^2 +(((√3)/2))^2 ))dt−(1/2)∫(√(m^2 +(((√3)/2))^2 ))dm       {suppose x−(1/2)=m  (t/4)(√(t^2 +(3/4)))+(3/(16))log(t+(√(t^2 +(3/4))))−(t/2)(√(m^2 +(3/4)))−(3/(16))log(m+(√(m^2 +(3/4))))  ((2x+1)/8)(√(x^2 +x+1))+(3/(16))log(x+(1/2)+(√(x^2 +x+1)))−((2x−1)/8)(√(x^2 −x+1))−(3/(16))log(x−(1/2)+(√(x^2 −x+1)))+C

$$\int\frac{{x}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\right)}{\mathrm{2}{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:\:}\left\{{suppose}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}={t}\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{t}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{m}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{dm}\:\:\:\:\:\:\:\left\{{suppose}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}={m}\right. \\ $$$$\frac{{t}}{\mathrm{4}}\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}+\frac{\mathrm{3}}{\mathrm{16}}{log}\left({t}+\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\right)−\frac{{t}}{\mathrm{2}}\sqrt{{m}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}−\frac{\mathrm{3}}{\mathrm{16}}{log}\left({m}+\sqrt{{m}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\right) \\ $$$$\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{8}}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{16}}{log}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)−\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{8}}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{16}}{log}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\right)+{C} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 02/Jul/20

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com