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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:… – Michael Joyce Oct 21 '12 at 22:46
$\frac12\log(1+\sqrt{2}x+x^2) + \frac12\log(1-\sqrt{2}x+x^2)$ is what you logarithm simplifies to. After that, use the chain rule. – Michael Hardy Oct 21 '12 at 23:00
There should be a minus sign there between the logs, Michael. – Aleks Vlasev Oct 22 '12 at 6:22

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


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. – pipi Oct 24 '12 at 7:17

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