Question Number 213216 by Spillover last updated on 01/Nov/24
Answered by MrGaster last updated on 01/Nov/24
![=∫_0 ^∞ (x^3 /((x+1)^2 (x^2 +1)(x^3 +1)))dx+∫_0 ^∞ ((log^2 (x))/((x+1)^2 (x^2 +1)(x^3 +1)))dx =∫_0 ^∞ (x^3 /((x+1)^2 (x^2 +1)(x^3 +1)))dx+[(∂^2 /∂a^2 )∫_0 ^∞ (x^(a−1) /((x+1)^2 (x^2 +1)(x^3 +1)))dx]_(a−1) =∫_0 ^∞ (x^3 /((x+1)^2 (x^2 +1)(x^3 +1)))dx+[(∂^2 /∂a^2 )((∂^2 /(sin(πa)))∙(((√3)−1)/( 4(√3))))]_(a=1) =∫_0 ^∞ (x^3 /((x+1)^2 (x^2 +1)(x^3 +1)))dx+(π^2 /6)∙(((√3)−1)/(4(√3))) =[(π/(4(√3)))−(π/(12))]+((π^2 ((√3)−1))/(24(√3)))](https://www.tinkutara.com/question/Q213225.png)
$$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{log}^{\mathrm{2}} \left({x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}+\left[\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}−\mathrm{1}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}\right]_{{a}−\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}+\left[\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\left(\frac{\partial^{\mathrm{2}} }{\mathrm{sin}\left(\pi{a}\right)}\centerdot\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\mathrm{4}\sqrt{\mathrm{3}}}\right)\right]_{{a}=\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\centerdot\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$=\left[\frac{\pi}{\mathrm{4}\sqrt{\mathrm{3}}}−\frac{\pi}{\mathrm{12}}\right]+\frac{\pi^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{24}\sqrt{\mathrm{3}}} \\ $$
Commented by Frix last updated on 01/Nov/24
$$\mathrm{Wrong}\:\mathrm{again}.\:\mathrm{You}\:\mathrm{get}\approx.\mathrm{365}\:\mathrm{but}\:\mathrm{it}\:\mathrm{should} \\ $$$$\mathrm{be}\:\approx\mathrm{1}.\mathrm{697} \\ $$