Question Number 56188 by maxmathsup by imad last updated on 11/Mar/19
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} }{{x}}\:{dx} \\ $$
Answered by Smail last updated on 12/Mar/19
![let t=((1+x))^(1/4) ⇒dx=4t^3 dt I=∫_0 ^∞ ((1/(x((1+x))^(1/4) ))−(1/(x(((1+x)^3 ))^(1/4) )))dx =4∫_1 ^∞ ((t^2 /(t^4 −1))−(1/(t^4 −1)))dt=4∫_1 ^∞ (((t^2 −1)/(t^4 −1)))dt =4∫_1 ^∞ (dt/(t^2 +1))=4[tan^(−1) (t)]_1 ^∞ =4((π/2)−(π/4)) =π](https://www.tinkutara.com/question/Q56263.png)
$${let}\:\:{t}=\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}}\Rightarrow{dx}=\mathrm{4}{t}^{\mathrm{3}} {dt} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}}}−\frac{\mathrm{1}}{{x}\sqrt[{\mathrm{4}}]{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }}\right){dx} \\ $$$$=\mathrm{4}\int_{\mathrm{1}} ^{\infty} \left(\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} −\mathrm{1}}−\frac{\mathrm{1}}{{t}^{\mathrm{4}} −\mathrm{1}}\right){dt}=\mathrm{4}\int_{\mathrm{1}} ^{\infty} \left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} −\mathrm{1}}\right){dt} \\ $$$$=\mathrm{4}\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{4}\left[{tan}^{−\mathrm{1}} \left({t}\right)\right]_{\mathrm{1}} ^{\infty} =\mathrm{4}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\pi \\ $$
Commented by maxmathsup by imad last updated on 12/Mar/19
$${sir}\:{smail}\:{do}\:{you}\:{study}\:{or}\:{work}\:{in}\:{usa}? \\ $$
Commented by Smail last updated on 12/Mar/19
$${I}\:{do}\:{both}\:{for}\:{now}. \\ $$