Hi there can anyone help me integrate the following
$\displaystyle \int $ (x^2+1)/(x^4+1)
I've been trying unsuccessfully
$\displaystyle I = \int\frac{x^2+1}{x^4+1}\;{dx} = \int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\;{dx } = \int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}\;{dx}$. Let t = x-1/x, then:
$\displaystyle I = \int\frac{1}{t^2+2}\;{dt} $ which hopefully is an integral that you are familiar with! :)
Yes (I'm full of ideas this morning, haha), note that (x^2+√2x+1)+(x^2-√2x+1) = 2(x^2+1), so that:
$\displaystyle \begin{aligned} I & = \int\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\;{dx} = \frac{1}{2}\int\frac{(x^2+\sqrt{2}x+1)+(x^2-\sqrt{2}x+1)}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\;{dx} \\& = \frac{1}{2}\int\frac{1}{x^2-\sqrt{2}x+1}\;{dx}+\frac{1}{2}\int\frac{1}{x^2+ \sqrt{2}x+1}\;{dx} = I_{1}+I_{2}, ~ \text{say}. \end{aligned}$
$\displaystyle I_{1} = \frac{1}{2}\int\frac{1}{\left(x-\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2 }}\right)^2}\;{dx} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x-1\right)+k_{1}$ and
$\displaystyle I_{2} = \frac{1}{2}\int\frac{1}{\left(x+\frac{1}{\sqrt{2}} \right)^2+\left(\frac{1}{\sqrt{2}}\right)^2}\;{dx} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x+1\right)+k_{2}$, thus:
$\displaystyle I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x-1\right)+ \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x+1\right)+k.$