Question Number 219509 by Nicholas666 last updated on 27/Apr/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{I}\:=\:\underset{\:\mathrm{1}} {\int}^{\:\mathrm{16}} \:\frac{\left({x}\:+\sqrt{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} }\:{dx} \\ $$$$ \\ $$
Answered by SdC355 last updated on 27/Apr/25
$$\int\:\centerdot=\frac{\mathrm{2}\centerdot^{\mathrm{4}} \sqrt{{x}^{\mathrm{3}} }\centerdot^{\mathrm{4}} \sqrt{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{3}} }}{\mathrm{3}\centerdot^{\mathrm{4}} \sqrt{\left({x}+\sqrt{{x}}\right)^{\mathrm{3}} }}\centerdot_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\:;\:\frac{\mathrm{7}}{\mathrm{4}};−\sqrt{{x}}\right)+\frac{\sqrt{{x}}+\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\left({x}+\sqrt{{x}}\right)^{\mathrm{3}} }} \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{16}} \:\centerdot=\frac{\mathrm{16}}{\mathrm{3}}\centerdot^{\mathrm{4}} \sqrt{\mathrm{2}}\left(\mathrm{5}\centerdot^{\mathrm{4}} \sqrt{\mathrm{10}}\centerdot_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\mathrm{1},\mathrm{2};\frac{\mathrm{7}}{\mathrm{4}};−\mathrm{4}\right)−_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\mathrm{1},\mathrm{2};\frac{\mathrm{7}}{\mathrm{4}};−\mathrm{1}\right)\right) \\ $$$$\mathrm{approximatly}\: \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{16}} \:\centerdot\approx\mathrm{6}.\mathrm{58781555261161938092791165}….. \\ $$
Commented by SdC355 last updated on 27/Apr/25
$$\:_{{p}} \boldsymbol{\mathrm{F}}_{{q}} \left({a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ….{a}_{{p}} ;{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ….{b}_{{q}} ;{z}\right)\:\mathrm{is}\:\mathrm{Hypergeometric}\:\mathrm{function} \\ $$
Commented by Nicholas666 last updated on 27/Apr/25
$${nice} \\ $$
Commented by Ghisom last updated on 28/Apr/25
$$\mathrm{please}\:\mathrm{show}\:\mathrm{the}\:\mathrm{steps}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{with}\:_{\mathrm{2}} {F}_{\mathrm{1}} .\:\mathrm{which}\:\mathrm{calculator}\:\mathrm{do}\:\mathrm{you}\:\mathrm{use}\:\mathrm{to} \\ $$$$\mathrm{get}\:\mathrm{27}\:\mathrm{digits}? \\ $$
Answered by Ghisom last updated on 27/Apr/25
![∫(((x+x^(1/2) )^(1/4) )/x^(3/4) )dx= [t=(((x^(1/2) +1)^(1/4) )/x^(1/8) ) → dx=−8x^(9/8) (x^(1/2) +1)^(3/4) dt] =−8∫(t^4 /((t^4 −1)^2 ))dt= [decompose etc] =((2t)/(t^4 −1))+(1/2)ln ∣((t+1)/(t−1))∣ +arctan t 1≤x≤16 ⇒ 2^(1/4) ≤t≤2^(−1/2) 5^(1/4) we could get an exact value for I but it′s not useful at all. I≈6.58781555261](https://www.tinkutara.com/question/Q219526.png)
$$\int\frac{\left({x}+{x}^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{1}/\mathrm{4}} }{{x}^{\mathrm{3}/\mathrm{4}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\left({x}^{\mathrm{1}/\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }{{x}^{\mathrm{1}/\mathrm{8}} }\:\rightarrow\:{dx}=−\mathrm{8}{x}^{\mathrm{9}/\mathrm{8}} \left({x}^{\mathrm{1}/\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{4}} {dt}\right] \\ $$$$=−\mathrm{8}\int\frac{{t}^{\mathrm{4}} }{\left({t}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{decompose}\:\mathrm{etc}\right] \\ $$$$=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\mid\:+\mathrm{arctan}\:{t} \\ $$$$\mathrm{1}\leqslant{x}\leqslant\mathrm{16}\:\Rightarrow\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} \leqslant{t}\leqslant\mathrm{2}^{−\mathrm{1}/\mathrm{2}} \mathrm{5}^{\mathrm{1}/\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{get}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{for}\:{I}\:\mathrm{but}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{not}\:\mathrm{useful}\:\mathrm{at}\:\mathrm{all}. \\ $$$${I}\approx\mathrm{6}.\mathrm{58781555261} \\ $$
Commented by Nicholas666 last updated on 28/Apr/25
$${thanks} \\ $$
Commented by Ghisom last updated on 28/Apr/25
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$