1. intergration

noting that conclude that

2. Originally Posted by rmpatel5
noting that conclude that
$f(x)=\frac{1+x^2}{1+x^4}$. So we can see that $f(-x)=f(x)$ so $f(x)$ is even. Thus

\begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}dx&=\fra c{1}{2}\cdot2\int_0^1\frac{1+x^2}{1+x^4}dx\\
&=\frac{1}{2}\int_{-1}^{1}\frac{1+x^2}{1+x^4}dx\end{aligned}

So using the given info we may rewrite this as

\begin{aligned}\frac{1}{2}\int_{-1}^{1}\frac{1+x^2}{1+x^4}dx&=\frac{1}{4}\int_{-1}^1\frac{2(1+x^2)}{1+x^4}dx\\
&=\frac{1}{4}\int_{-1}^1\bigg[\frac{1}{1+\sqrt{2}x+x^2}+\frac{1}{1-\sqrt{2}x+x^2}\bigg]
\end{aligned}

Now note that $\frac{1}{1+\sqrt{2}(-x)+x^2}=\frac{1}{1-\sqrt{2}x+x^2}$

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\\
&=\frac{1}{2}\int_{-1}^{1}\frac{dx}{1+\sqrt{2}x+x^2}dx\\
&=\frac{1}{2}\int_{-1}^1\frac{dx}{1-\sqrt{2}x+x^2}dx\\
&=\int_0^1\frac{1+x^2}{1+x^4}dx
\end{aligned}