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This problem is not solved.

$$ \begin{align} f(x) &=\log\ \sqrt{\frac{1+\sqrt{2}x +x^2}{1-\sqrt{2}x +x^2}}+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right) \cr \frac{df}{dx}&=\mathord? \end{align} $$

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    For the log term, you'll want to use properties of logarithms to simply *before* you differentiate. There's also probably a clever way to use the arctangent addition formula found here: http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Arctangent_addition_formula2012-10-21
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    $\frac12\log(1+\sqrt{2}x+x^2) + \frac12\log(1-\sqrt{2}x+x^2)$ is what you logarithm simplifies to. After that, use the chain rule.2012-10-21
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    There should be a minus sign there between the logs, Michael.2012-10-22

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The answer is $$\frac{2\sqrt{2}}{1+x^4}$$

Hints: $$\frac{d\left(\log(1+x^2\pm\sqrt{2}x\right)}{dx}=\frac{2x\pm\sqrt{2}}{1+x^2\pm\sqrt{2}x}$$ and $$\left(1+x^2+\sqrt{2}x\right)\left(1+x^2-\sqrt{2}x\right)=\left(1+x^2\right)^2-\left(\sqrt{2}x\right)^2$$
Similarly, $$\frac{d\left(\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right)\right)}{dx}=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{d\left(\frac{\sqrt{2}x}{1-x^2}\right)}{dx}$$

$$=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{\sqrt{2}}{2}\left(\frac{1}{\left(1-x\right)^2}+\frac{1}{\left(1+x\right)^2}\right)$$

The rest is by calculations.

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    To get the derivation is not difficult, but tedious. However if the question asks for $\int\frac{1}{1+x^4}dx$, then it is more complicated.2012-10-24