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Question-139328




Question Number 139328 by Bekzod Jumayev last updated on 25/Apr/21
Commented by Bekzod Jumayev last updated on 25/Apr/21
Help please?
$$\boldsymbol{{Help}}\:\boldsymbol{{please}}? \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 26/Apr/21
∫_0 ^∞ (dx/( (√(x^3 +x))))=∫_0 ^∞ (dx/( (√x)∙(√(x^2 +1)))) ,x=tanθ ⇒dx=sec^2 θdθ  =∫_0 ^(π/2) ((sec^2 θdθ)/( (√(tanθ))∙(√(sec^2 θ))))=∫_0 ^(π/2) ((secθ)/( (√(tanθ))))dθ  =∫_0 ^(π/2) sin^(−(1/2)) θcos^(−(1/2)) θdθ=(1/2)β((1/4),(1/4))=((Γ^2 ((1/4)))/(2Γ((1/2))))  =(1/(2(√π)))Γ^2 ((1/4))
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{x}}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}}\centerdot\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\:,\mathrm{x}=\mathrm{tan}\theta\:\Rightarrow\mathrm{dx}=\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta}{\:\sqrt{\mathrm{tan}\theta}\centerdot\sqrt{\mathrm{sec}^{\mathrm{2}} \theta}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}\theta}{\:\sqrt{\mathrm{tan}\theta}}\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\pi}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 26/Apr/21
∫_0 ^∞ (dx/( (√(x^3 +x))))        x^2 =u  =(1/2)∫_0 ^∞ (u^((1/4)−1) /((1+u)^((1/4)+(1/4)) ))du=(1/2)((Γ^2 ((1/4)))/(Γ((1/2))))=((Γ^2 ((1/4)))/( 2(√π)))
$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} +{x}}}\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}} }{du}=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\:\mathrm{2}\sqrt{\pi}} \\ $$

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