Hi there can anyone help me integrate the following

$\displaystyle \int $ (x^2+1)/(x^4+1)

I've been trying unsuccessfully :(

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- Apr 28th 2011, 09:48 PMuserodIntegration
Hi there can anyone help me integrate the following

$\displaystyle \int $ (x^2+1)/(x^4+1)

I've been trying unsuccessfully :( - Apr 28th 2011, 09:54 PMpickslides
What have you tried?

Maybe break it up using partial fractions? - Apr 28th 2011, 10:13 PMTheCoffeeMachine
$\displaystyle I = \int\frac{x^2+1}{x^4+1}\;{dx} = \int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\;{dx } = \int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}\;{dx}$. 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! :) - Apr 28th 2011, 10:14 PMuserod
Ya that's what I've been trying I've $\displaystyle x^4+1$ factored as follows

(x^2+(2^1/2)x+1)(x^2-(2^1/2)x+1)

I'm having trouble solving the partial fractions though. Any ideas? - Apr 28th 2011, 10:33 PMProve It
x^4 + 1 doesn't factorise... You should follow TheCoffeeMachine's method.

- Apr 28th 2011, 10:46 PMuserod
Great thanks ya I know that one inverse tan over root 2 all over root 2 right??!!! Thanks again guys really helpful site!

- Apr 28th 2011, 10:59 PMTheCoffeeMachine
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:

$\displaystyle I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x-1\right)+ \frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}x+1\right)+k.$ - Apr 29th 2011, 03:39 AMmr fantastic