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Question Number 42790 by maxmathsup by imad last updated on 02/Sep/18
1) calculate  ∫_0 ^∞   (t/(1+t^4 )) dt  2) calculate ∫_0 ^1    (t/(1+t^4 ))dt  3) calculste ∫_1 ^(+∞)   (t/(1+t^4 ))dt
$$\left.\mathrm{1}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculste}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\: \\ $$
Answered by MJS last updated on 03/Sep/18
∫(t/(t^4 +1))dt=(1/2)∫(du/(u^2 +1))=(1/2)arctan u=  =(1/2)arctan t^2  +C  (1)=(π/4)  (2)=(3)=(π/8)
$$\int\frac{{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{u}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{t}^{\mathrm{2}} \:+{C} \\ $$$$\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)=\left(\mathrm{3}\right)=\frac{\pi}{\mathrm{8}} \\ $$

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