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Originally Posted by rmpatel5
noting that. So we can see that
so
is even. Thus
So using the given info we may rewrite this as
Now note that
So they are refelctions of each other over the y-axis, so while they may not be equal, they have the same area underneath.
so
![\begin{aligned}\frac{1}{4}\int_{-1}^1\bigg[\frac{1}{1+\sqrt{2}x+x^2}+\frac{1}{1-\sqrt{2}x+x^2}\bigg]dx&=\frac{1}{4}\int_{-1}^{1}\frac{2}{1+\sqrt{2}x+x^2}dx\\<br />
&=\frac{1}{2}\int_{-1}^{1}\frac{dx}{1+\sqrt{2}x+x^2}dx\\<br />
&=\frac{1}{2}\int_{-1}^1\frac{dx}{1-\sqrt{2}x+x^2}dx\\<br />
&=\int_0^1\frac{1+x^2}{1+x^4}dx<br />
\end{aligned}](http://latex.codecogs.com/png.latex?\begin{aligned}\frac{1}{4}\int_{-1}^1\bigg[\frac{1}{1+\sqrt{2}x+x^2}+\frac{1}{1-\sqrt{2}x+x^2}\bigg]dx&=\frac{1}{4}\int_{-1}^{1}\frac{2}{1+\sqrt{2}x+x^2}dx\\<br />
&=\frac{1}{2}\int_{-1}^{1}\frac{dx}{1+\sqrt{2}x+x^2}dx\\<br />
&=\frac{1}{2}\int_{-1}^1\frac{dx}{1-\sqrt{2}x+x^2}dx\\<br />
&=\int_0^1\frac{1+x^2}{1+x^4}dx<br />
\end{aligned})
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