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Question Number 168732 by mnjuly1970 last updated on 16/Apr/22

	Answered by mnjuly1970 last updated on 16/Apr/22
	$Ω = ∫_0 ^( ∞) (( x^( (1/4)) + 2x^( (5/4)) +x^( (9/4)) )/((1+x^( 3) )^( 2) ))dx =^(x^( 3) = t) (1/3)∫_0 ^( ∞) ((t^( (1/(12))−(8/(12))) + 2t^( (5/(12))−(8/(12))) + t^( (9/(12))−(8/(12))) )/((1 + t )^( 2) )) dt = (1/3) ∫_0 ^( ∞) ((t^( ((−7)/(12))) +2t^( ((−1)/4)) + t^( ((−1)/(12))) )/((1+ t )^( 2) ))dt = (1/(3 )) ∫_0 ^( ∞) ((t^( (5/(12)) −1) + 2t^( (3/4) −1) + t^( ((11)/(12))−1) )/((1 + t )^( 2) ))dt = (1/3) { β ((5/(12)) , ((19)/(12))) +2β((3/4) , (5/4)) +β (((11)/(12)) , ((13)/(12)) } = (1/3) { ((Γ((5/(12)) )Γ( ((19)/(12)))+2Γ((3/4))Γ((5/4))+Γ (((11)/(12)))Γ (((13)/(12)) ) )/([Γ(2)= 1 ]))} = (1/3) { (7/(12)) (π/(sin(((5π)/(12))))) +(1/2) (π/(sin((π/4)))) +(1/(12)) (π/(sin((π/(12))))) } = (1/3) { (7/(3 )) (π/( (√6) +(√2))) + (π/( (√2))) +(1/3) (π/( (√6) −(√2))) } = (π/3) { ((7((√6) −(√2) ))/(12)) +(1/( (√2))) + (((√6) +(√2))/(12)) } = (π/3) { ((8(√6) −6(√2) )/(12)) + (1/( (√2) )) } = (π/3)(((8(√6))/(12)))= (2/3) ((√6)/3)π = (2/3)(√(2/3)) π ∴ Ω = (√(4/9)) ((((4/9))))^(1/4) π a − 1 = (4/9) ⇒ a = ((13)/9) ✓$
	$Ω = \int_{0}^{\infty} \frac{x^{\frac{1}{4}} + 2 x^{\frac{5}{4}} + x^{\frac{9}{4}}}{{(1 + x^{3})}^{2}} d x$ $\overset{x^{3} = t}{=} \frac{1}{3} \int_{0}^{\infty} \frac{t^{\frac{1}{12} - \frac{8}{12}} + 2 t^{\frac{5}{12} - \frac{8}{12}} + t^{\frac{9}{12} - \frac{8}{12}}}{{(1 + t)}^{2}} d t$ $= \frac{1}{3} \int_{0}^{\infty} \frac{t^{\frac{- 7}{12}} + 2 t^{\frac{- 1}{4}} + t^{\frac{- 1}{12}}}{{(1 + t)}^{2}} d t$ $= \frac{1}{3} \int_{0}^{\infty} \frac{t^{\frac{5}{12} - 1} + 2 t^{\frac{3}{4} - 1} + t^{\frac{11}{12} - 1}}{{(1 + t)}^{2}} d t$ $= \frac{1}{3} {β (\frac{5}{12}, \frac{19}{12}) + 2 β (\frac{3}{4}, \frac{5}{4}) + β (\frac{11}{12}, \frac{13}{12}}$ $= \frac{1}{3} {\frac{Γ (\frac{5}{12}) Γ (\frac{19}{12}) + 2 Γ (\frac{3}{4}) Γ (\frac{5}{4}) + Γ (\frac{11}{12}) Γ (\frac{13}{12})}{[Γ (2) = 1]}}$ $= \frac{1}{3} {\frac{7}{12} \frac{π}{s i n (\frac{5 π}{12})} + \frac{1}{2} \frac{π}{s i n (\frac{π}{4})} + \frac{1}{12} \frac{π}{s i n (\frac{π}{12})}}$ $= \frac{1}{3} {\frac{7}{3} \frac{π}{\sqrt{6} + \sqrt{2}} + \frac{π}{\sqrt{2}} + \frac{1}{3} \frac{π}{\sqrt{6} - \sqrt{2}}}$ $= \frac{π}{3} {\frac{7 (\sqrt{6} - \sqrt{2})}{12} + \frac{1}{\sqrt{2}} + \frac{\sqrt{6} + \sqrt{2}}{12}}$ $= \frac{π}{3} {\frac{8 \sqrt{6} - 6 \sqrt{2}}{12} + \frac{1}{\sqrt{2}}}$ $= \frac{π}{3} (\frac{8 \sqrt{6}}{12}) = \frac{2}{3} \frac{\sqrt{6}}{3} π = \frac{2}{3} \sqrt{\frac{2}{3}} π$ $∴ Ω = \sqrt{\frac{4}{9}} \sqrt[4]{(\frac{4}{9})} π$ $a - 1 = \frac{4}{9} \Rightarrow a = \frac{13}{9} ✓$
	Commented by safojontoshtemirov last updated on 16/Apr/22
	very nice ��
	Commented by Tawa11 last updated on 16/Apr/22

	$Great sir .$
	Answered by Mathspace last updated on 17/Apr/22

	$Ψ = \int_{0}^{\infty} \frac{x^{- \frac{1}{4}}}{{(x^{2} - x + 1)}^{2}} d x \Rightarrow$ $Ψ = \int_{0}^{\infty} \frac{x^{- \frac{1}{4}}}{{(\frac{x^{3} + 1}{x + 1})}^{2}} d x$ $= \int_{0}^{\infty} \frac{x^{- \frac{1}{4}} (x^{2} + 2 x + 1)}{{(1 + x^{3})}^{2}} d x$ $= \int_{0}^{\infty} \frac{x^{\frac{7}{4}} + 2 x^{\frac{3}{4}} + x^{- \frac{1}{4}}}{{(1 + x^{3})}^{2}} d x$ $l e t φ (a, b) = \int_{0}^{\infty} \frac{x^{a - 1}}{x^{3} + b^{3}} d x \Rightarrow$ $\frac{\partial φ}{\partial b} = - 3 b^{2} \int_{0}^{\infty} \frac{x^{a - 1}}{{(x^{3} + b^{3})}^{2}} d x \Rightarrow$ $\int_{0}^{\infty} \frac{x^{a - 1}}{{(x^{3} + 1)}^{2}} d x = - \frac{1}{3} \frac{\partial φ}{\partial b} ∣_{b = 1}$ $φ (a, b) =_{x = b z} \int_{0}^{\infty} \frac{{(b z)}^{a - 1}}{b^{3} (1 + z^{3})} b d z$ $= b^{a - 3} \int_{0}^{\infty} \frac{z^{a - 1}}{1 + z^{3}} d z (z^{3} = t)$ $= b^{a - 3} \int_{0}^{\infty} \frac{t^{\frac{a - 1}{3}}}{1 + t} \frac{1}{3} t^{\frac{1}{3} - 1} d t$ $= \frac{b^{a - 3}}{3} \int_{0}^{\infty} \frac{t^{\frac{a}{3} - 1}}{1 + t} d t$ $= \frac{b^{a - 3}}{3} \times \frac{π}{s i n (\frac{a π}{3})} \Rightarrow$ $\frac{\partial φ}{\partial b} (a, b) = \frac{(a - 3}{3} b^{a - 4} \frac{π}{s i n (\frac{a π}{3})} \Rightarrow$ $\frac{\partial φ}{\partial b} ∣_{b = 1} = \frac{a - 3}{3} \times \frac{π}{s i n (\frac{π a}{3})}$ $\Rightarrow \int_{0}^{\infty} \frac{x^{a - 1}}{{(x^{3} + 1)}^{2}} d x = \frac{π (3 - a)}{9 s i n (\frac{π a}{3})}$ $\Rightarrow \int_{0}^{\infty} \frac{x^{\frac{7}{4}}}{{(1 + x^{3})}^{2}} d x = \int_{0}^{\infty} \frac{x^{\frac{11}{4} - 1}}{{(1 + x^{3})}^{2}} d x$ $= \frac{π (3 - \frac{11}{4})}{9 s i n (\frac{11 π}{12})} = . . .$ $\int_{0}^{\infty} \frac{x^{\frac{3}{4}}}{{(1 + x^{3})}^{2}} d x = \int_{0}^{\infty} \frac{x^{\frac{7}{4} - 1}}{{(1 + x^{3})}^{2}} d x$ $= \frac{π (3 - \frac{7}{4})}{9 s i n (\frac{7 π}{12})} = . . .$
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