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Question Number 53470 by maxmathsup by imad last updated on 22/Jan/19

$$findVn={\int }_{\frac{1}{n}}^{\frac{an-1}{n}}\frac{\sqrt{x}}{\sqrt{a-\sqrt{x}+x}}dx$$$$$$

Commented by maxmathsup by imad last updated on 24/Jan/19

$$changement\sqrt{x}=tgivex=t^2\Rightarrow dx=2tdtand$$$$V_n={\int }_{\frac{1}{\sqrt{n}}}^{\sqrt{\frac{an-1}{n}}}\frac{t}{\sqrt{a-t+t^2}}\left(2t\right)dt=2{\int }_{\frac{1}{\sqrt{n}}}^{\sqrt{\frac{an-1}{n}}}\frac{t^2}{\sqrt{t^2-2\frac{1}{2}t+\frac{1}{4}+a-\frac{1}{4}}}dt$$$$=2{\int }_{\frac{1}{\sqrt{n}}}^{\sqrt{\frac{an-1}{n}}}\frac{t^2}{\sqrt{{(t-\frac{1}{2})}^2+\frac{4a-1}{4}}}letsupposea>\frac{1}{4}\Rightarrow$$$$V_n{=}_{t-\frac{1}{2}=\frac{\sqrt{4a-1}}{2}u}2{\int }_{\frac{\frac{2}{\sqrt{n}}-1}{\sqrt{4a-1}}}^{\frac{2\sqrt{\frac{an-1}{n}}-1}{\sqrt{4a-1}}}\frac{{(\frac{\sqrt{4a-1}}{2}u+\frac{1}{2})}^2}{\frac{\sqrt{4a-1}}{2}\sqrt{1+u^2}}\frac{\sqrt{4a-1}}{2}du$$$$=2{\int }_{\frac{2-\sqrt{n}}{\sqrt{n}\sqrt{4a-1}}}^{\frac{2(\sqrt{4a-1}-\sqrt{n}}{\sqrt{n}\sqrt{4a-1}}}\frac{1}{4}\frac{(4a-1)u^2+2\sqrt{4a-1}u+1}{\sqrt{1+u^2}}du$$$$=\frac{1}{2}{\int }_{{\alpha }_n}^{{\beta }_n}\frac{(4a-1)u^2+2\sqrt{4a-1}u+1}{\sqrt{1+u^2}}du$$$$=\frac{4a-1}{2}{\int }_{{\alpha }_n}^{{\beta }_n}\frac{u^2+1-1}{\sqrt{1+u^2}}+\sqrt{4a-1}{\int }_{{\alpha }_n}^{{\beta }_n}\frac{u}{\sqrt{1+u^2}}du+\frac{1}{2}{\int }_{{\alpha }_n}^{{\beta }_n}\frac{du}{\sqrt{1+u^2}}$$$${\int }_{{\alpha }_n}^{{\beta }_n}\frac{du}{\sqrt{1+u^2}}=[ln(1+\sqrt{{1+x^2]}_{{\alpha }_n}^{{\beta }_n}}=ln(1+\sqrt{1+{\beta }_n^2})-ln(1+\sqrt{1+{\alpha }_n^2})$$$${\int }_{{\alpha }_n}^{{\beta }_n}\frac{u}{\sqrt{1+u^2}}du={\left[\sqrt{1+u^2}\right]}_{{\alpha }_n}^{{\beta }_n}=\sqrt{1+{\beta }_n^2}-\sqrt{1+{\alpha }_n^2}$$$${\int }_{{\alpha }_n}^{{\beta }_n}\frac{u^2+1-1}{\sqrt{1+u^2}}du={\int }_{{\alpha }_n}^{{\beta }_n}\sqrt{1+u^2}du-{\int }_{{\alpha }_n}^{{\beta }_n}\frac{du}{\sqrt{1+u^2}}=.....$$$$$$$$$$$$$$

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