$\int_{-5}^{5} \log \left(\frac{7-x}{7+x}\right) d x=$

Question

Aicharya · Accepted Answer

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

$\int_{-5}^{5} \log \left(\frac{7-x}{7+x}\right) d x=$

Solution