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$\int_{-5}^{5} \log \left(\frac{7-x}{7+x}\right) d x=$
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Let $1=\int_{-4}^{5} \log \frac{7-x}{7+x}$
Let $\quad f(x)=\log \frac{7-x}{7+x}$
$f(-x)=\log \left[\frac{7-(-x)}{7+(-x)}\right]=\log \left(\frac{7+x}{7-x}\right)=-\log \left(\frac{7-x}{7+x}\right)=-f(x)$
$\therefore f(x)$ is an odd function $\Rightarrow I=0$
Let $1=\int_{-4}^{5} \log \frac{7-x}{7+x}$
Let $\quad f(x)=\log \frac{7-x}{7+x}$
$f(-x)=\log \left[\frac{7-(-x)}{7+(-x)}\right]=\log \left(\frac{7+x}{7-x}\right)=-\log \left(\frac{7-x}{7+x}\right)=-f(x)$
$\therefore f(x)$ is an odd function $\Rightarrow I=0$
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