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Question Number 164609 by SANOGO last updated on 19/Jan/22

$$enposantx=t-\frac{1}{t}$$$${\underset{0}{\int }}^{+oo}\frac{1+t^2}{1+t^4}dt$$

Answered by Mathspace last updated on 19/Jan/22

$$I={\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt={\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+2}dt$$$${=}_{t-\frac{1}{t}=x}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+2}$$$${=}_{x=\sqrt{2}y}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sqrt{2}dy}{2(1+y^2)}$$$$=\frac{\sqrt{2}}{2}{\left[arctan\left(y\right)\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}$$$$=\frac{\sqrt{2}}{2}(\frac{\pi }{2}+\frac{\pi }{2})=\frac{\pi \sqrt{2}}{2}$$

Commented by SANOGO last updated on 19/Jan/22

$$mercibien$$

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