Question Number 53470 by maxmathsup by imad last updated on 22/Jan/19
$${find}\:\:{Vn}=\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{{an}−\mathrm{1}}{{n}}} \:\frac{\sqrt{{x}}}{\:\sqrt{{a}−\sqrt{{x}}+{x}}}{dx} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 24/Jan/19
![changement (√x)=t give x=t^2 ⇒dx =2tdt and V_n = ∫_(1/( (√n))) ^(√((an−1)/n)) (t/( (√(a−t+t^2 )))) (2t)dt =2 ∫_(1/( (√n))) ^(√((an−1)/n)) (t^2 /( (√(t^2 −2(1/2)t +(1/4)+a−(1/4)))))dt =2 ∫_(1/( (√n))) ^(√((an−1)/n)) (t^2 /( (√((t−(1/2))^2 +((4a−1)/4))))) let suppose a>(1/4) ⇒ V_n =_(t−(1/2)=((√(4a−1))/2)u) 2∫_(((2/( (√n)))−1)/( (√(4a−1)))) ^((2(√((an−1)/n))−1)/( (√(4a−1)))) (((((√(4a−1))/2)u+(1/2))^2 )/(((√(4a−1))/2)(√(1+u^2 )))) ((√(4a−1))/2) du = 2 ∫_((2−(√n))/( (√n)(√(4a−1)))) ^((2((√(4a−1))−(√n))/( (√n)(√(4a−1)))) (1/4) (((4a−1)u^2 +2(√(4a−1))u +1)/( (√(1+u^2 )))) du =(1/2) ∫_α_n ^β_n (((4a−1)u^2 +2(√(4a−1))u +1)/( (√(1+u^2 )))) du =((4a−1)/2) ∫_α_n ^β_n ((u^2 +1 −1)/( (√(1+u^2 )))) + (√(4a−1))∫_α_n ^β_n (u/( (√(1+u^2 )))) du +(1/2) ∫_α_n ^β_n (du/( (√(1+u^2 )))) ∫_α_n ^β_n (du/( (√(1+u^2 )))) =[ln(1+(√(1+x^2 ]_α_n ^β_n ))=ln(1+(√(1+β_n ^2 )))−ln(1+(√(1+α_n ^2 ))) ∫_α_n ^β_n (u/( (√(1+u^2 )))) du =[(√(1+u^2 ))]_α_n ^β_n =(√(1+β_n ^2 ))−(√(1+α_n ^2 )) ∫_α_n ^β_n ((u^2 +1−1)/( (√(1+u^2 )))) du =∫_α_n ^β_n (√(1+u^2 ))du −∫_α_n ^β_n (du/( (√(1+u^2 )))) =.....](https://www.tinkutara.com/question/Q53674.png)
$${changement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow{dx}\:=\mathrm{2}{tdt}\:{and} \\ $$$${V}_{{n}} =\:\int_{\frac{\mathrm{1}}{\:\sqrt{{n}}}} ^{\sqrt{\frac{{an}−\mathrm{1}}{{n}}}} \:\:\frac{{t}}{\:\sqrt{{a}−{t}+{t}^{\mathrm{2}} }}\:\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int_{\frac{\mathrm{1}}{\:\sqrt{{n}}}} ^{\sqrt{\frac{{an}−\mathrm{1}}{{n}}}} \:\:\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}}+{a}−\frac{\mathrm{1}}{\mathrm{4}}}}{dt} \\ $$$$=\mathrm{2}\:\int_{\frac{\mathrm{1}}{\:\sqrt{{n}}}} ^{\sqrt{\frac{{an}−\mathrm{1}}{{n}}}} \:\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}{a}−\mathrm{1}}{\mathrm{4}}}}\:\:\:{let}\:{suppose}\:{a}>\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${V}_{{n}} =_{{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}{u}} \:\:\:\:\:\:\mathrm{2}\int_{\frac{\frac{\mathrm{2}}{\:\sqrt{{n}}}−\mathrm{1}}{\:\sqrt{\mathrm{4}{a}−\mathrm{1}}}} ^{\frac{\mathrm{2}\sqrt{\frac{{an}−\mathrm{1}}{{n}}}−\mathrm{1}}{\:\sqrt{\mathrm{4}{a}−\mathrm{1}}}} \:\:\:\frac{\left(\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}{u}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\:{du} \\ $$$$=\:\mathrm{2}\:\int_{\frac{\mathrm{2}−\sqrt{{n}}}{\:\sqrt{{n}}\sqrt{\mathrm{4}{a}−\mathrm{1}}}} ^{\frac{\mathrm{2}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}−\sqrt{{n}}\right.}{\:\sqrt{{n}}\sqrt{\mathrm{4}{a}−\mathrm{1}}}} \:\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\left(\mathrm{4}{a}−\mathrm{1}\right){u}^{\mathrm{2}} \:+\mathrm{2}\sqrt{\mathrm{4}{a}−\mathrm{1}}{u}\:+\mathrm{1}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\:\frac{\left(\mathrm{4}{a}−\mathrm{1}\right){u}^{\mathrm{2}} \:+\mathrm{2}\sqrt{\mathrm{4}{a}−\mathrm{1}}{u}\:+\mathrm{1}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{du} \\ $$$$=\frac{\mathrm{4}{a}−\mathrm{1}}{\mathrm{2}}\:\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\:\frac{{u}^{\mathrm{2}} \:+\mathrm{1}\:−\mathrm{1}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:+\:\sqrt{\mathrm{4}{a}−\mathrm{1}}\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\frac{{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{du}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:=\left[{ln}\left(\mathrm{1}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right]_{\alpha_{{n}} } ^{\beta_{{n}} } }={ln}\left(\mathrm{1}+\sqrt{\mathrm{1}+\beta_{{n}} ^{\mathrm{2}} }\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{1}+\alpha_{{n}} ^{\mathrm{2}} }\right)\right.\right. \\ $$$$\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\:\frac{{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{du}\:=\left[\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right]_{\alpha_{{n}} } ^{\beta_{{n}} } \:=\sqrt{\mathrm{1}+\beta_{{n}} ^{\mathrm{2}} }−\sqrt{\mathrm{1}+\alpha_{{n}} ^{\mathrm{2}} } \\ $$$$\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\:\frac{{u}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{du}\:=\int_{\alpha_{{n}} } ^{\beta_{{n}} } \sqrt{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:−\int_{\alpha_{{n}} } ^{\beta_{{n}} } \:\:\frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:=….. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$