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Question Number 203681 by SANOGO last updated on 25/Jan/24

$${\int }_1^{+oo}\frac{1}{x\sqrt{x^2+x+1}}dx$$

Answered by Frix last updated on 25/Jan/24

$$t=\frac{2x+1+2\sqrt{x^2+x+1}}{\sqrt{3}}$$$$\Rightarrow$$$$\underset{2+\sqrt{3}}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{t-\sqrt{3}}-\frac{3}{3t+\sqrt{3}}dt={\left[\ln \frac{t-\sqrt{3}}{3t+\sqrt{3}}\right]}_{2+\sqrt{3}}^{\mathrm{\infty }}=$$$$=\ln (3+2\sqrt{3})-\ln 3$$

Answered by esmaeil last updated on 26/Jan/24

$${\underset{1}{\int }}^{+\mathrm{\infty }}\frac{dx}{x\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}=$$$$$$$$\text{Missing left or extra right}$$$$\frac{2x+1}{\sqrt{3}}=tan\theta \to dx=\frac{\sqrt{3}}{2{cos}^2\theta }d\theta \to$$$$\mathrm{\Omega }=\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{2}{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{\frac{d\theta }{{cos}^2\theta }}{tan\theta \times \frac{1}{vos\theta }}=$$$$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}cosec\theta d\theta =$$$${ln\mid cosec\theta -cot\theta \mid ]}_{\frac{\pi }{3}}^{\frac{\pi }{2}}=$$$$ln\mid (1-\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{3})\mid =$$$$ln(\sqrt{3}-1)$$$$m$$

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