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Question Number 30773 by abdo imad last updated on 25/Feb/18

$$leta>0calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2+a^2)}^2}$$ $$2)calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2}{{(x^2+a^2)}^2}dx.$$

Commented byabdo imad last updated on 26/Feb/18

$$1)thech.x=atantgive$$ $$I={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2+a^2)}^2}={\int }_0^{\frac{\pi }{2}}\frac{1}{a^4{(1+{tan}^{2}t)}^2}a(1+{tan}^{2}t)dt$$ $$=\frac{1}{a^3}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{tan}^{2}t}\Rightarrow a^{3}I={\int }_0^{\frac{\pi }{2}}{cos}^{2}tdt$$ $$={\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(2t\right)}{2}dt=\frac{\pi }{4}+{\left[\frac{1}{4}sin\left(2t\right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{4}\Rightarrow$$ $$I=\frac{\pi }{4a^3}$$ $$2)letputJ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2}{{(x^2+a^2)}^2}dx$$ $$J=2{\int }_0^{\mathrm{\infty }}\frac{x^2+a^2-a^2}{{(x^2+a^2)}^2}dx=2{\int }_0^{\mathrm{\infty }}\frac{dx}{(x^2+a^2)}dx-2a^2{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2+a^2)}^2}dx$$ $$chx=at\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{(x^2+a^2)}={\int }_0^{\mathrm{\infty }}\frac{adt}{a^2(1+t^2)}=\frac{1}{a}\frac{\pi }{2}=\frac{\pi }{2a}$$ $$J=\frac{\pi }{a}-2a^{2}I=\frac{\pi }{a}-2a^2\frac{\pi }{4a^3}=\frac{\pi }{a}-\frac{\pi }{2a}\Rightarrow J=\frac{\pi }{2a}.$$

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