Find the value of $ \int_0^{\infty} \frac{2x^2-1}{4x^4+1}\,{dx} $ 
Find $ \displaystyle \int_0^{\infty} \frac{2x^2-1}{4x^4+1}\,{dx}
 $

I can find the primitive, but the fact that the answer is $0$ made me suspect there might be a way of getting answer without finding the primitive. I tried letting $x = 1/y$ but it didn't get me anywhere. 

Instead of substituting $x = \dfrac{1}{y}$, try substituting $x = \dfrac{1}{2y}$. 
This gives you: 
\begin{align*}I &= \displaystyle\int_{0}^{\infty}\dfrac{2x^2-1}{4x^4+1}\,dx \\ &= \displaystyle\int_{\infty}^{0}\dfrac{\tfrac{1}{2y^2}-1}{\tfrac{1}{4y^4}+1} \cdot -\dfrac{1}{2y^2}\,dy \\ &= \displaystyle\int_{0}^{\infty}\dfrac{\tfrac{1}{2y^2}-1}{\tfrac{1}{4y^4}+1} \cdot \dfrac{2y^2}{4y^4}\,dy \\ &= \displaystyle\int_{0}^{\infty}\dfrac{1-2y^2}{4y^4+1}\,dy \\ &= -\displaystyle\int_{0}^{\infty}\dfrac{2y^2-1}{4y^4+1}\,dy \\ &= -I.\end{align*}
Since $I = -I$, we have $I = 0$.
