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Question-148884




Question Number 148884 by 0731619 last updated on 01/Aug/21
Answered by aleks041103 last updated on 03/Aug/21
∫_0 ^∞ (dx/(a+x^(13) ))=(1/a)∫_0 ^∞ (dx/(1+((x/a^(1/13) ))^(13) ))=  =(a^(1/13) /a)∫_0 ^∞ ((d(x/a^(1/13) ))/(1+(x/a^(1/13) )^(13) ))  if a>0, then we substitute   u=x/a^(1/13)   Then:  ∫_0 ^∞ (dx/(a+x^(13) ))=a^(−12/13) ∫_0 ^∞ (dx/(1+x^(13) ))  ⇒((∫_0 ^∞ (dx/(1+x^(13) )))/(∫_0 ^∞ (dx/(13+x^(13) ))))=13^(12/13)
$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{a}+{x}^{\mathrm{13}} }=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+\left(\frac{{x}}{{a}^{\mathrm{1}/\mathrm{13}} }\right)^{\mathrm{13}} }= \\ $$$$=\frac{{a}^{\mathrm{1}/\mathrm{13}} }{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{d}\left({x}/{a}^{\mathrm{1}/\mathrm{13}} \right)}{\mathrm{1}+\left({x}/{a}^{\mathrm{1}/\mathrm{13}} \right)^{\mathrm{13}} } \\ $$$${if}\:{a}>\mathrm{0},\:{then}\:{we}\:{substitute}\: \\ $$$${u}={x}/{a}^{\mathrm{1}/\mathrm{13}} \\ $$$${Then}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{a}+{x}^{\mathrm{13}} }={a}^{−\mathrm{12}/\mathrm{13}} \int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{13}} } \\ $$$$\Rightarrow\frac{\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{13}} }}{\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{13}+{x}^{\mathrm{13}} }}=\mathrm{13}^{\mathrm{12}/\mathrm{13}} \\ $$

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