Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

$${\int }_0^1\frac{dx}{\sqrt{x^2+x+1}}=?$$

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

$$letI={\int }_0^1\frac{dx}{\sqrt{x^2+x+1}}\Rightarrow I={\int }_0^1\frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}$$$$cha7gementx+\frac{1}{2}=\frac{\sqrt{3}}{2}tan\theta give\theta =arctan\left(\frac{2x+1}{\sqrt{3}}\right)$$$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{\sqrt{\frac{3}{4}}\sqrt{1+{tan}^2\theta }}\times \frac{\sqrt{3}}{2}(1+{tan}^2\theta )d\theta ={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{1+{tan}^2\theta }d\theta$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{d\theta }{cos\theta }{=}_{tan\left(\frac{\theta }{2}\right)=u}{\int }_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}}\frac{1}{\frac{1-u^2}{1+u^2}}\times \frac{2du}{1+u^2}$$$$={\int }_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}}(\frac{1}{1-u}+\frac{1}{1+u})du={[ln\mid \frac{1+u}{1-u}\mid ]}_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}}$$$$=ln\mid \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\mid -ln\mid \frac{3-\sqrt{3}}{-1+\sqrt{3}}\mid =ln\mid \sqrt{3}+1\mid -ln\mid \sqrt{3}-1\mid -ln\mid \frac{\sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1}\mid$$$$=ln(\sqrt{3}+1)-ln(\sqrt{3}-1)-ln\left(\sqrt{3}\right)$$

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

$$LetnameditA$$$$using1+x+x^2={(x+\frac{1}{2})}^2+\frac{3}{4}=\frac{3}{4}[{\left(\frac{2x+1}{\sqrt{3}}\right)}^2+1]$$$$$$$$A={\int }_0^1\frac{2}{\sqrt{3}}\frac{1}{\sqrt{[{\left(\frac{2x+1}{\sqrt{3}}\right)}^2+1}}dx$$$$={\left[argsh\left(\frac{2x+1}{\sqrt{3}}\right)\right]}_0^1=argsh\left(\sqrt{3}\right)-argsh\left(\frac{1}{\sqrt{3}}\right)$$$$usingargshx=ln(x+\sqrt{1+x^2})$$$$A=ln(2+\sqrt{3})-ln\left(\sqrt{3}\right)$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com