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Question Number 35139 by behi83417@gmail.com last updated on 15/May/18

Commented by abdo mathsup 649 cc last updated on 16/May/18

$$letputI={\int }_0^{\sqrt{2}}\frac{x+1}{\sqrt{x^2+x+1}}dx$$$$I=\frac{1}{2}{\int }_0^{\sqrt{2}}\frac{2x+2}{\sqrt{x^2+x+1}}dx$$$$=\frac{1}{2}{\int }_0^{\sqrt{2}}\frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{1}{2}{\int }_0^{\sqrt{2}}\frac{dx}{\sqrt{x^2+x+1}}but$$$${\int }_0^{\sqrt{2}}\frac{2x+1}{\sqrt{x^2+x+1}}dx={\left[2\sqrt{x^2+x+1}\right]}_0^{\sqrt{2}}$$$$=2\sqrt{3+\sqrt{2}}-2wehave{x}^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}$$$$andchangementx+\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(t\right)givesh\left(t\right)=\frac{2x+1}{\sqrt{3}}$$$${\int }_0^{\sqrt{2}}\frac{dx}{\sqrt{x^2+x+1}}={\int }_{argsh\left(\frac{1}{\sqrt{3}}\right)}^{argsh\left(\frac{2\sqrt{2}+1}{\sqrt{3}}\right)}\frac{1}{\sqrt{\frac{3}{4}(1+{sh}^{2}t)}}\frac{\sqrt{3}}{2}chtdt$$$$={\int }_{argsh\left(\frac{1}{\sqrt{3}}\right)}^{argsh\left(\frac{2\sqrt{2}+1}{\sqrt{3}}\right)}dt=argsh\left(\frac{2\sqrt{2}+1}{\sqrt{3}}\right)-argsh\left(\frac{1}{\sqrt{3}}\right)$$$$=ln(\frac{2\sqrt{2}+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2\sqrt{2}+1}{\sqrt{3}}\right)}^2})$$$$-ln(\frac{1}{\sqrt{3}}+\sqrt{1+{\left(\frac{1}{\sqrt{3}}\right)}^2})so$$$$I=\sqrt{3+\sqrt{2}}-1+\frac{1}{2}ln(\frac{2\sqrt{2}+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2\sqrt{2}+1}{\sqrt{3}}\right)}^2})$$$$-ln(\frac{1}{\sqrt{3}}+\sqrt{1+{\left(\frac{1}{\sqrt{3}}\right)}^2).}$$

Commented by abdo mathsup 649 cc last updated on 16/May/18

$$letputx+1=t\Rightarrow$$$$I={\int }_1^{1+\sqrt{2}}\frac{t}{\sqrt{{(t-1)}^2+t}}dt$$$$={\int }_1^{1+\sqrt{2}}\frac{t}{\sqrt{t^2-t+1}}dt$$$$={\int }_1^{1+\sqrt{2}}\frac{t}{\sqrt{{(t-\frac{1}{2})}^2+\frac{3}{4}}}dtchangement$$$${(t-\frac{1}{2})}^{}=\frac{\sqrt{3}}{2}u\Rightarrow u=\frac{2t-1}{\sqrt{3}}$$$$I={\int }_{\frac{1}{\sqrt{3}}}^{\frac{1+2\sqrt{2}}{\sqrt{3}}}\frac{1+u\sqrt{3}}{2\frac{\sqrt{3}}{2}\sqrt{u^2+1}}du$$$$I=\frac{1}{\sqrt{3}}{\int }_{\frac{1}{\sqrt{3}}}^{\frac{1+2\sqrt{2}}{\sqrt{3}}}\frac{du}{\sqrt{1+u^2}}+{\int }_{\frac{1}{\sqrt{3}}}^{\frac{1+2\sqrt{2}}{\sqrt{3}}}\frac{u}{\sqrt{u^2+1}}du$$$$I=\frac{1}{\sqrt{3}}[ln{(u+\sqrt{1+u^2}]}_{\frac{1}{\sqrt{3}}}^{\frac{1+2\sqrt{2}}{\sqrt{3}}}+{\left[\sqrt{1+u^2}\right]}_{\frac{1}{\sqrt{3}}}^{\frac{1+2\sqrt{2}}{\sqrt{3}}}so$$$$thevalueofIisdetermined....$$

Answered by MJS last updated on 16/May/18

$$\int \frac{x+1}{\sqrt{x^2+x+1}}dx=\int \frac{\frac{1}{2}(2x+1)+\frac{1}{2}}{\sqrt{x^2+x+1}}dx=$$$$=\frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}=$$$$$$$$\frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx=$$$$[t=x^2+x+1\to dx=\frac{dt}{2x+1}]$$$$=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\frac{1}{2}2\sqrt{t}=\sqrt{x^2+x+1}$$$$$$$$\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}=\frac{1}{2}\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}=$$$$[u=\frac{\sqrt{3}}{3}(2x+1)\to dx=\frac{\sqrt{3}}{2}du]$$$$=\frac{1}{2}\int \frac{du}{\sqrt{u^2+1}}=\frac{1}{2}\ln \mid \sqrt{u^2+1}+u\mid =$$$$=\frac{1}{2}\ln \mid \sqrt{{\left(\frac{\sqrt{3}}{3}(2x+1)\right)}^2+1}+\frac{\sqrt{3}}{2}(2x+1)\mid =$$$$=\frac{1}{2}\ln \mid 2\sqrt{x^2+x+1}+2x+1\mid -\frac{\ln \left(3\right)}{4}$$$$$$$$=\sqrt{x^2+x+1}+\frac{1}{2}\ln \mid 2\sqrt{x^2+x+1}+2x+1\mid +C$$$$$$$$\underset{0}{\stackrel{\sqrt{2}}{\int }}\frac{x+1}{\sqrt{x^2+x+1}}dx=\sqrt{3+\sqrt{2}}+\frac{1}{2}\ln (2\sqrt{3+\sqrt{2}}+2\sqrt{2}+1)-(1+\frac{\ln \left(3\right)}{2})=$$$$=\sqrt{3+\sqrt{2}}-1+\frac{1}{2}\ln \left(\frac{2\sqrt{3+\sqrt{2}}+2\sqrt{2}+1}{3}\right)\approx 1.593316$$

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