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Question-201573




Question Number 201573 by sonukgindia last updated on 09/Dec/23
Answered by witcher3 last updated on 09/Dec/23
=∫_(−∞) ^∞ ((e^(−2024x) +e^(−2020) )/((e^(−2025x) +e^(−2019x) )((−4x^3 +(4x^2 +1)(√(1+x^2 ))−3x)^(2023) +1)))=I  =∫_(−∞) ^∞ ((e^x +e^(5x) )/((1+e^(6x) )((−4x^3 −3x+(4x^2 +1)(√(x^2 +1)))(((−4x^3 −3x−(4x^2 +1)(√(x^2 +1)))/(−4x^3 −3x−(4x^2 +1)(√(x^2 +1))))))^2 +1)))  {(4x^3 +3x)+(4x^2 +1)(√(x^2 +1))}{.(4x^3 +3x)−(4x^2 +1)(√(x^2 +1}))=−1  I=∫_(−∞) ^∞ (((e^x +e^(5x) )(4x^3 +3x+(4x^2 +1)(√(x^2 +1)))^(2023) )/((1+e^(6x) )(1+(4x^3 +3x+(4x^2 +1)(√(x^2 +1)))^(2023) ))dx  2I=∫_(−∞) ^∞ ((e^x +e^(5x) )/(1+e^(6x) ))dx  =∫_0 ^∞ ((1+x^4 )/(1+x^6 ))dx,x^6 =y  =∫_0 ^∞ ((1+y^(2/3) )/(1+y)).(y^(−(5/6)) /6)dy  =(1/6)∫_0 ^∞ (y^(−(5/6)) /(1+y))dy+(1/6)∫_0 ^∞ (y^(−(1/6)) /(1+y))dy  =(1/6)β((1/6),(5/6))+(1/6)β((5/6),(1/6))=(1/3).(π/(sin((π/6))))  =((2π)/3)  I=(π/3)
$$=\int_{−\infty} ^{\infty} \frac{\mathrm{e}^{−\mathrm{2024x}} +\mathrm{e}^{−\mathrm{2020}} }{\left(\mathrm{e}^{−\mathrm{2025x}} +\mathrm{e}^{−\mathrm{2019x}} \right)\left(\left(−\mathrm{4x}^{\mathrm{3}} +\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{3x}\right)^{\mathrm{2023}} +\mathrm{1}\right)}=\mathrm{I} \\ $$$$=\int_{−\infty} ^{\infty} \frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{\mathrm{5x}} }{\left.\left(\mathrm{1}+\mathrm{e}^{\mathrm{6x}} \right)\left(\left(−\mathrm{4x}^{\mathrm{3}} −\mathrm{3x}+\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(\frac{−\mathrm{4x}^{\mathrm{3}} −\mathrm{3x}−\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}{−\mathrm{4x}^{\mathrm{3}} −\mathrm{3x}−\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\right)\right)^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\left\{\left(\mathrm{4x}^{\mathrm{3}} +\mathrm{3x}\right)+\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right\}\left\{.\left(\mathrm{4x}^{\mathrm{3}} +\mathrm{3x}\right)−\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\left.\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right\}}=−\mathrm{1}\right. \\ $$$$\mathrm{I}=\int_{−\infty} ^{\infty} \frac{\left(\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{\mathrm{5x}} \right)\left(\mathrm{4x}^{\mathrm{3}} +\mathrm{3x}+\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2023}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{6x}} \right)\left(\mathrm{1}+\left(\mathrm{4x}^{\mathrm{3}} +\mathrm{3x}+\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2023}} \right.}\mathrm{dx} \\ $$$$\mathrm{2I}=\int_{−\infty} ^{\infty} \frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{\mathrm{5x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{6x}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx},\mathrm{x}^{\mathrm{6}} =\mathrm{y} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\mathrm{y}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}+\mathrm{y}}.\frac{\mathrm{y}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\mathrm{6}}\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\mathrm{1}+\mathrm{y}}\mathrm{dy}+\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{−\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{1}+\mathrm{y}}\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\beta\left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{5}}{\mathrm{6}}\right)+\frac{\mathrm{1}}{\mathrm{6}}\beta\left(\frac{\mathrm{5}}{\mathrm{6}},\frac{\mathrm{1}}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{I}=\frac{\pi}{\mathrm{3}} \\ $$$$ \\ $$
Commented by Anonim_X last updated on 09/Dec/23
nice
Commented by Calculusboy last updated on 09/Dec/23
nice solution
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$
Commented by witcher3 last updated on 10/Dec/23
thank You
$$\mathrm{thank}\:\mathrm{You} \\ $$

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