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Question Number 186263 by mokys last updated on 02/Feb/23

$${\int }_0^1\sqrt{1+\frac{1}{4x}}dx$$$$$$

Answered by cortano1 last updated on 03/Feb/23

$$let\frac{1}{2\sqrt{x}}=\tan t\Rightarrow 2\sqrt{x}=\cot t$$$$\Rightarrow 4x=\cot {}^{2}t$$$$\Rightarrow 4dx=2\cot t(-{\csc }^{2}t)dt$$$$I=\int \frac{1}{\sqrt{1+\tan {}^{2}t}}(-2\cot t{\csc }^{2}t)dt$$$$I=-2\int \cos t\frac{\cos t}{\sin t}.\frac{1}{\sin {}^{2}t}dt$$$$I=-2\int \frac{1-\sin {}^{2}t}{\sin {}^{3}t}dt$$$$I=-2\int ({\csc }^{3}t-\csc t)dt$$$$$$

Answered by Mathspace last updated on 03/Feb/23

$$let\frac{1}{4x}=t\Rightarrow x=\frac{1}{4t}\Rightarrow$$$$I=-{\int }_{\frac{1}{4}}^{+\mathrm{\infty }}\sqrt{1+t}\left(\frac{-dt}{4t^2}\right)$$$$4I={\int }_{\frac{1}{4}}^{+\mathrm{\infty }}\frac{\sqrt{1+t}}{t^2}dt(1+t=z^2)$$$$={\int }_{\frac{\sqrt{5}}{2}}^{\mathrm{\infty }}\frac{z}{{(z^2-1)}^2}\left(2z\right)dz$$$$={\int }_{\frac{\sqrt{5}}{2}}z\left(\frac{2z}{{(z^2-1)}^2}\right)dz\left(byparts\right)$$$$={[-\frac{z}{z^2-1}]}_{\frac{\sqrt{5}}{2}}^{\mathrm{\infty }}-{\int }_{\frac{\sqrt{5}}{2}}^{\mathrm{\infty }}(-\frac{1}{z^2-1})dz$$$$=\frac{\sqrt{5}}{2(\frac{5}{4}-1)}+\frac{1}{2}{\int }_{\frac{\sqrt{5}}{2}}^{\mathrm{\infty }}(\frac{1}{z-1}-\frac{1}{z+1})dz$$$$=2\sqrt{5}+\frac{1}{2}{[ln\mid \frac{z-1}{z+1}\mid ]}_{\frac{\sqrt{5}}{2}}^{\mathrm{\infty }}$$$$=2\sqrt{5}-\frac{1}{2}ln\mid \frac{\frac{\sqrt{5}}{2}-1}{\frac{\sqrt{5}}{2}+1}\mid$$$$=2\sqrt{5}-\frac{1}{2}ln\left(\frac{\sqrt{5}-2}{\sqrt{5}+2}\right)$$

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