Question

If $I_1={\int }_0^{\infty }\frac{dx}{1+x^4}$ and $I_2={\int }_0^{\infty }\frac{x^2}{1+x^4}dx$, then

• A. $I_1=I_2$
• B. $I_1=2I_2$
• C. $2I_1=I_2$
• D. none of these

Answer 1

The correct option is A $I_1=I_2$

$I_2={\int }_0^{\infty }\frac{x^2}{1+x^4}dx={\int }_0^{\infty }\frac{\frac{1}{x^2}}{1+\frac{1}{x^4}}dx$

Put $t=\frac{1}{x}⟹dt=-\frac{dx}{x^2}$

$I_2={\int }_{\infty }^0\frac{-dt}{1+t^4}={\int }_0^{\infty }\frac{dx}{1+x^4}=I_1$