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### Answers (1)

Answer: $\frac{4}{1+x^{2}+x^{4}}$

Hint: To solve this equation we use $\left ( \frac{u}{v} \right )^{'}$  form

Given:    $y=\frac{\log \left(x^{2}+x+1\right)}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$

Solution:

$\frac{d y}{d x}=\left(\frac{(2 x+1)\left(x^{2}-x+1\right)-(2 x-1)\left(x^{2}+x+1\right.}{\left(x^{2}-x+1\right)^{2}}\right)+\frac{2}{\sqrt{3}}\left(\frac{1}{\frac{1+3 x^{2}}{\left(1-x^{2}\right)^{2}}}\right)\left(\frac{\sqrt{3}\left(1-x^{2}\right)-\sqrt{3} x(-2 x)}{\left(1-x^{2}\right)^{2}}\right)$

$=\frac{2 x^{3}+x^{2}-2 x^{2}-x+2 x+1-2 x^{3}-2 x^{2}-2 x+x^{2}+x+1}{\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)}\; \; +$   $\frac{2}{\sqrt{x}} \cdot \frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}+x^{4}\right)}\left[\frac{\sqrt{3}-\sqrt{3} x^{2}+2 \sqrt{3} x^{2}}{\left(1-x^{2}\right)^{2}}\right]$

\begin{aligned} &=\frac{2-2 x^{2}+2+2 x^{2}}{1+x^{2}+x^{4}} \\\\ &=\frac{4}{1+x^{2}+x^{4}} \end{aligned}

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