Question Number 87086 by Chi Mes Try last updated on 02/Apr/20
Answered by mind is power last updated on 03/Apr/20
$$=\int_{\mathrm{0}} ^{+\infty} \mathrm{2}\frac{{dt}}{\left({t}^{\mathrm{4}} +\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} +{t}^{\mathrm{4}} −………+{t}^{\mathrm{100}} \right)}={A} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{dt}}{{t}^{\mathrm{104}} \left(\mathrm{1}+\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right)\left(\frac{\mathrm{1}}{{t}^{\mathrm{100}} }−\frac{\mathrm{1}}{{t}^{\mathrm{98}} }+…….+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{{t}}={y}\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{\mathrm{102}} }{\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){y}^{\mathrm{2}} +{y}^{\mathrm{4}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} +………+{y}^{\mathrm{100}} \right)} \\ $$$$\mathrm{1}−{y}^{\mathrm{2}} +…….+{y}^{\mathrm{100}} =\frac{\mathrm{1}−\left(−{y}^{\mathrm{2}} \right)^{\mathrm{51}} }{\mathrm{1}+{y}^{\mathrm{2}} }=\frac{\mathrm{1}+{y}^{\mathrm{102}} }{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$$\mathrm{2}{A}=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{2}+\mathrm{2}{x}^{\mathrm{102}} {dx}}{\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} −…..+{x}^{\mathrm{100}} \right)} \\ $$$$\Rightarrow{A}=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}.\frac{\mathrm{1}+{x}^{\mathrm{102}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} +……..+{x}^{\mathrm{100}} \right)}{dx} \\ $$$${A}=\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}\:\: \\ $$$${easy}\:{now} \\ $$$$ \\ $$