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Question Number 156218 by talminator2856791 last updated on 09/Oct/21
∫_0 ^( 1) (1/( (√(x(√(x^2 (√(x^3 (√(x^4 +1)))))))) )) dx
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{x}\sqrt{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}}}\:}\:{dx} \\ $$$$\: \\ $$
Commented by talminator2856791 last updated on 09/Oct/21
who can solve this monster?
$$\:\mathrm{who}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{monster}?\: \\ $$
Commented by MJS_new last updated on 09/Oct/21
thank you for making it clear yoz're only posting questions for fun. I won't have to care anymore. but one last thing: 1+1=2 now go see if wolfram alpha gives the same answer. in this case I'm guilty of copying. or your logic is faulty.
Commented by MJS_new last updated on 10/Oct/21
you mean somebody pays for wolfram alpha only to answer questions on this forum? the free version only gives results without the path...
Commented by talminator2856791 last updated on 10/Oct/21
your logic is faulty as 1+1=2 is not equivalent to monster integral when you use same variables and techniques as wolframalpha, it is very questionable.
$$\:\mathrm{your}\:\mathrm{logic}\:\mathrm{is}\:\mathrm{faulty}\:\mathrm{as}\:\mathrm{1}+\mathrm{1}=\mathrm{2}\:\: \\ $$$$\:\mathrm{is}\:\mathrm{not}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{monster}\:\mathrm{integral}\: \\ $$$$\:\mathrm{when}\:\mathrm{you}\:\mathrm{use}\:\mathrm{same}\:\mathrm{variables}\:\mathrm{and}\:\: \\ $$$$\:\mathrm{techniques}\:\mathrm{as}\:\mathrm{wolframalpha},\:\: \\ $$$$\:\mathrm{it}\:\mathrm{is}\:\mathrm{very}\:\mathrm{questionable}.\:\: \\ $$
Answered by mindispower last updated on 09/Oct/21
=∫_0 ^1 (√x).(√(√x^2 )).(√((√((√x^3 ). )).))(1+x^4 )^(1/(16)) =∫_0 ^1 x^((1/2)+(1/2)+(3/8)) (1+x^4 )^(1/(16)) =∫_0 ^1 x^((11)/8) (1+x^4 )^(1/(16)) =∫_0 ^1 (1/4)t^(((11)/(32))−(3/4)) (1+t)^(1/(16)) dt =(1/4)∫_0 ^1 t^((−13)/(32)) (1+t)^(1/(16)) dt ∫_0 ^1 t^x (1+t)^y dt=,−1<x ,g(x,y) =∫_0 ^1 (1+Σ_(n≥1) Π_(k=0) ^(n−1) (((y−k))/(n!))t^n )t^x dt =(1/(x+1))+Σ_(n≥1) ∫_0 ^1 (−1)^n Π_(k=0) ^(n−1) (((k−y))/(n!))t^(n+x) dt =(1/(1+x))+Σ_(n≥1) (−1)^n ((Π_(k=0) ^(n−1) (−y+k))/(n!.(n+x+1))) n+x+1=((Γ(n+x+2))/(Γ(n+x+1))) =(1/(1+x))+Σ_(n≥1) (((−1)^n Π_k ^(n−1) (−y+k)(x+1+k))/((1+x)n!.Π_(k=0) ^(n−1) (x+2+k))) =(1/(1+x))+(1/(1+x))Σ_(n≥1) (((−y)_n (1+x)_n )/((x+2)_n )).(((−1)^n )/(n!)) =(1/(1+x))(1+Σ_(n≥1) (((−y)_n (1+x)_n )/((2+x)_n )).(((−1)^n )/(n!))) =(1/(1+x))(_2 F_1 (−y,1+x;2+x;−1) x=((−13)/(32)),y=(1/(16)) We get ∫_0 ^1 (√(x(√(x^2 (√(x^3 (√(1+x^4 ))))))))dx=((32)/(19)) _2 F_1 (−(1/(16)),((19)/(32));((51)/(32)),−1)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}}.\sqrt{\sqrt{{x}^{\mathrm{2}} }}.\sqrt{\sqrt{\sqrt{{x}^{\mathrm{3}} }.\:}.}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{8}}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{11}}{\mathrm{8}}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{\mathrm{11}}{\mathrm{32}}−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{16}}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{−\mathrm{13}}{\mathrm{32}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{16}}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{x}} \left(\mathrm{1}+{t}\right)^{{y}} {dt}=,−\mathrm{1}<{x}\:,{g}\left({x},{y}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left({y}−{k}\right)}{{n}!}{t}^{{n}} \right){t}^{{x}} {dt} \\ $$$$=\frac{\mathrm{1}}{{x}+\mathrm{1}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left({k}−{y}\right)}{{n}!}{t}^{{n}+{x}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(−{y}+{k}\right)}{{n}!.\left({n}+{x}+\mathrm{1}\right)} \\ $$$${n}+{x}+\mathrm{1}=\frac{\Gamma\left({n}+{x}+\mathrm{2}\right)}{\Gamma\left({n}+{x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \underset{{k}} {\overset{{n}−\mathrm{1}} {\prod}}\left(−{y}+{k}\right)\left({x}+\mathrm{1}+{k}\right)}{\left(\mathrm{1}+{x}\right){n}!.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}+\mathrm{2}+{k}\right)}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−{y}\right)_{{n}} \left(\mathrm{1}+{x}\right)_{{n}} }{\left({x}+\mathrm{2}\right)_{{n}} }.\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−{y}\right)_{{n}} \left(\mathrm{1}+{x}\right)_{{n}} }{\left(\mathrm{2}+{x}\right)_{{n}} }.\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{x}}\left(_{\mathrm{2}} {F}_{\mathrm{1}} \left(−{y},\mathrm{1}+{x};\mathrm{2}+{x};−\mathrm{1}\right)\right. \\ $$$${x}=\frac{−\mathrm{13}}{\mathrm{32}},{y}=\frac{\mathrm{1}}{\mathrm{16}}\:{We}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}\sqrt{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}}}{dx}=\frac{\mathrm{32}}{\mathrm{19}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{16}},\frac{\mathrm{19}}{\mathrm{32}};\frac{\mathrm{51}}{\mathrm{32}},−\mathrm{1}\right) \\ $$
Commented by talminator2856791 last updated on 09/Oct/21
all the answers he post is same as the wolframalpha.
$$\:\mathrm{all}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{he}\:\mathrm{post}\:\mathrm{is}\:\mathrm{same}\:\mathrm{as}\:\: \\ $$$$\:\mathrm{the}\:\mathrm{wolframalpha}. \\ $$
Commented by talminator2856791 last updated on 10/Oct/21
Mr W i dont want to examine people i do this for fun
$$\:\mathrm{Mr}\:\mathrm{W}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{want}\:\mathrm{to}\:\mathrm{examine}\:\mathrm{people}\:\: \\ $$$$\:\mathrm{i}\:\mathrm{do}\:\mathrm{this}\:\mathrm{for}\:\mathrm{fun}\:\: \\ $$
Commented by talminator2856791 last updated on 09/Oct/21
i delete my comment because i know Mr W will look for a fight. and i dont want to fight.
$$\:\mathrm{i}\:\mathrm{delete}\:\mathrm{my}\:\mathrm{comment}\:\mathrm{because}\:\mathrm{i}\:\mathrm{know}\:\: \\ $$$$\:\mathrm{Mr}\:\mathrm{W}\:\mathrm{will}\:\mathrm{look}\:\mathrm{for}\:\mathrm{a}\:\mathrm{fight}. \\ $$$$\:\mathrm{and}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{want}\:\mathrm{to}\:\mathrm{fight}. \\ $$
Commented by peter frank last updated on 09/Oct/21
true sir Mr W.i think the only way do not help him
$$\mathrm{true}\:\mathrm{sir}\:\mathrm{Mr}\:\mathrm{W}.\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{only}\:\mathrm{way} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{help}\:\mathrm{him}\: \\ $$
Commented by peter frank last updated on 09/Oct/21
To talminator.you have to show some respect
$$\mathrm{To}\:\mathrm{talminator}.\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{some}\:\mathrm{respect} \\ $$
Commented by talminator2856791 last updated on 10/Oct/21
you think posting wolframlpha answers is respect?
$$\:\mathrm{you}\:\mathrm{think}\:\mathrm{posting}\:\mathrm{wolframlpha}\:\: \\ $$$$\:\mathrm{answers}\:\mathrm{is}\:\mathrm{respect}?\:\:\: \\ $$
Commented by talminator2856791 last updated on 10/Oct/21
Mr W why you delete your post now? mr complain you questioned me and i responded.
$$\:\mathrm{Mr}\:\mathrm{W}\:\mathrm{why}\:\mathrm{you}\:\mathrm{delete}\:\mathrm{your}\:\mathrm{post}\:\mathrm{now}?\:\: \\ $$$$\:\mathrm{mr}\:\mathrm{complain}\:\: \\ $$$$\:\mathrm{you}\:\mathrm{questioned}\:\mathrm{me}\:\mathrm{and}\:\mathrm{i}\:\mathrm{responded}.\:\: \\ $$
Commented by mr W last updated on 10/Oct/21
i deleted my comments basically because i don′t want to have anything to do with people like you. i said you are posting questions here not because you want to learn or you are interested in ways how to solve. in fact you are posting the questions to “examine” other people. actually you have admited that what i said. so it′s clear for me now and i won′t spend any my time for it. p.s.: i have never wanted to fight anybody, you neither. what i requested is solely respect to other people. that you called me “mr complain” shows that you absolutely don′t know what is respect.
$${i}\:{deleted}\:{my}\:{comments}\:{basically} \\ $$$${because}\:{i}\:{don}'{t}\:{want}\:{to}\:{have}\:{anything} \\ $$$${to}\:{do}\:{with}\:{people}\:{like}\:{you}.\:{i}\:{said} \\ $$$${you}\:{are}\:{posting}\:{questions}\:{here}\:{not} \\ $$$${because}\:{you}\:{want}\:{to}\:{learn}\:{or}\:{you}\:{are} \\ $$$${interested}\:{in}\:{ways}\:{how}\:{to}\:{solve}.\:{in} \\ $$$${fact}\:{you}\:{are}\:{posting}\:{the}\:{questions}\: \\ $$$${to}\:“{examine}''\:{other}\:{people}.\:{actually} \\ $$$${you}\:{have}\:{admited}\:{that}\:{what}\:{i}\:{said}. \\ $$$${so}\:{it}'{s}\:{clear}\:{for}\:{me}\:{now}\:{and}\:\:{i}\:{won}'{t}\: \\ $$$${spend}\:{any}\:{my}\:{time}\:{for}\:{it}. \\ $$$${p}.{s}.:\:{i}\:{have}\:{never}\:{wanted}\:{to}\:{fight} \\ $$$${anybody},\:{you}\:{neither}.\:{what}\:{i} \\ $$$${requested}\:\:{is}\:\:{solely}\:{respect}\: \\ $$$${to}\:{other}\:{people}.\:{that}\:{you}\:{called}\:{me} \\ $$$$“{mr}\:{complain}''\:{shows}\:{that}\:{you}\: \\ $$$${absolutely}\:{don}'{t}\:{know}\:{what}\:{is}\:{respect}. \\ $$

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