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$\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} d x$ is equal to
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$I=\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} d x$
Let $f(x)=\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$
Replacing $x$ by $-x$, we get
$\begin{aligned}
& f(-x)=\frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}} \\
& =\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}=-f(x)
\end{aligned}$
So, $f(x)$ is an odd function.
$\therefore \quad \int_{-1}^1 f(x) d x=0$
Let $f(x)=\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$
Replacing $x$ by $-x$, we get
$\begin{aligned}
& f(-x)=\frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}} \\
& =\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}=-f(x)
\end{aligned}$
So, $f(x)$ is an odd function.
$\therefore \quad \int_{-1}^1 f(x) d x=0$
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