# Evaluate:$\int_{-1}^1 log \bigg(\large\frac{4-x}{4+x}\bigg)dx$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$\int _{-a}^a f(x) dx =0$ if f(x) is an odd function
• (iii)If $f(-x)= -f(x),$ then it is an odd function
Given $I= \int \limits_{-1}^1 log \frac{(4-x)}{(4+x)}dx$

Let $f(x)=log \frac{(4-x)}{(4+x)}$ then $f(x)=log \bigg(\frac{4+x}{4-x}\bigg)$

If x is replaced by -x

then $f(-x)=log \bigg(\frac{4+x}{4-x} \bigg)=-f(x)$

Hence it is an odd function

$\int \limits_{-a}^a f(x) =0$

Therefore $I=0$

answered Mar 15, 2013 by