<s>[INST] Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
 [/INST] let $I=\int \dfrac{dx}{1+x^4}$
I=$\dfrac {1}{2}\int \dfrac {x^2+1-(x^2-1)}{1+x^4} $
splitting the fraction,
I=$\dfrac {1}{2} (\int \dfrac {x^2+1}{1+x^4} - \int \dfrac {x^2-1}{1+x^4} )$ 
let $I_1$ =$ \int \frac {x^2+1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_1$= $ \int \dfrac {1+\dfrac {1}{x^2} }{x^2 +\dfrac {1}{x^2} } $
put $x-\dfrac{1}{x}=t  $ for $I_1$ (observe that the derivative of $x-\dfrac{1}{x}=t  $ is $1+\dfrac{1}{x^2}$)
Let $I_2 = \int \dfrac {x^2-1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_2 = \int \dfrac {1-\dfrac{1}{x^2}}{x^2+\dfrac {1}{x^2}}$
put $x+\dfrac{1}{x}=t  $ for $I_2$ (observe that the derivative of $x+\dfrac{1}{x}=t  $ is $1-\dfrac{1}{x^2}$)
So, $ I=\dfrac {1}{2} (I_1-I_2) $
Finally, substitute the limits.
Hope this helps.. Cheers!
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