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Question Number 42809 by maxmathsup by imad last updated on 02/Sep/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{tdt}{{(1+t^4)}^2}$$

Commented by prof Abdo imad last updated on 03/Sep/18

$$letA={\int }_0^{\mathrm{\infty }}\frac{tdt}{{(1+t^4)}^2}changement{t}^2=xgive$$$$2tdt=dx\Rightarrow A={\int }_0^{\mathrm{\infty }}\frac{dx}{2{(1+x^2)}^2}$$$${=}_{x=tan\theta }\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^2\theta }{{(1+{tan}^2\theta )}^2}d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{cos}^2\theta d\theta$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+cos\left(2\theta \right))d\theta =\frac{\pi }{8}+\frac{1}{8}{\left[sin\left(2\theta \right)\right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{8}+0\Rightarrow A=\frac{\pi }{8}.$$

Answered by sma3l2996 last updated on 03/Sep/18

$$A={\int }_0^{\mathrm{\infty }}\frac{tdt}{{(1+t^4)}^2}$$$$t^2=x\Rightarrow dx=2tdt$$$$A=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^2)}^2}$$$$byparts$$$$u=\frac{1}{{(1+x^2)}^2}\Rightarrow u^{\prime }=\frac{-2\times 2x}{(1+x^2)}$$$$v=1\Rightarrow v^{\prime }=x$$$$A=\frac{1}{2}{\left[\frac{x}{{(1+x^2)}^2}\right]}_0^{\mathrm{\infty }}+2{\int }_0^{\mathrm{\infty }}\frac{x^2}{1+x^2}dx=2{\int }_0^{\mathrm{\infty }}(1-\frac{1}{1+x^2})dx$$$$A=2{[x-{tan}^{-1}\left(x\right)]}_0^{\mathrm{\infty }}$$$$A=+\mathrm{\infty }$$

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