intergration

Quote:

Originally Posted by rmpatel5

noting that http://qaboard.cramster.com/Answer-B...8612507539.gifconclude that http://qaboard.cramster.com/Answer-B...7362504823.gif

$f(x)=\frac{1+x^2}{1+x^4}$ . 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

$\begin{aligned}\frac{1}{4}\int_{-1}^1\bigg[\frac{1}{1+\sqrt{2}x+x^2}+\frac{1}{1-\sqrt{2}x+x^2}\bigg]dx&=\frac{1}{4}\int_{-1}^{1}\frac{2}{1+\sqrt{2}x+x^2}dx\\ &=\frac{1}{2}\int_{-1}^{1}\frac{dx}{1+\sqrt{2}x+x^2}dx\\ &=\frac{1}{2}\int_{-1}^1\frac{dx}{1-\sqrt{2}x+x^2}dx\\ &=\int_0^1\frac{1+x^2}{1+x^4}dx \end{aligned}$