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Question Number 183559 by cortano1 last updated on 27/Dec/22

$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x^2+\sqrt{x}}=?$$

Answered by greougoury555 last updated on 27/Dec/22

$$G={\int }_1^{\mathrm{\infty }}\frac{dx}{x^2+\sqrt{x}}={\int }_1^{\mathrm{\infty }}\frac{2du}{u^3+1}$$$$=\frac{1}{3}\left[{\int }_1^{\mathrm{\infty }}(\frac{2}{u+1}-\frac{2u-4}{u^2-u+1})du\right]$$$$=\frac{1}{3}[2\ln \mid u+1\mid -{\int }_1^{\mathrm{\infty }}\frac{2(u-\frac{1}{2})-3}{{(u-\frac{1}{2})}^2+\frac{3}{4}}du]$$$$=\frac{1}{3}{[2\ln \mid u+1\mid -\ln ({(u-\frac{1}{2})}^2+\frac{3}{4})+2\sqrt{3}{\tan }^{-1}\left(\frac{2(u-\frac{1}{2})}{\sqrt{3}}\right)]}_1^{\mathrm{\infty }}$$$$=\frac{1}{3}{[\ln \mid \frac{{(u+1)}^2}{u^2-u+1}\mid +2\sqrt{3}{\tan }^{-1}\left(\frac{2u-1}{\sqrt{3}}\right)]}_1^{\mathrm{\infty }}$$$$=\frac{2\pi }{3\sqrt{3}}-\ln \sqrt[3]{4}$$$$=\frac{2\pi \sqrt{3}}{9}-\ln \sqrt[3]{4}$$

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