Question

Please help me with question 2

Answer 1

$I= \int_{-1}^{1} \log \bigg({\frac{3-x}{3+x}\bigg)} dx \: \: \: \: (1)$


There is a well known property of definite integration which is extremely useful here,

$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$

We will use this property in the given Integral.

Therefore,

$I = \int_{-1}^{1} \log\bigg(\frac{3-(-1+1-x)}{3+(-1+1-x)}\bigg) dx \\ I=\int_{-1}^{1} \log\bigg(\frac{3+x}{3-x}\bigg) dx \: \: \: \: \:(2)$
Add (1) and (2)


$2I = \int_{-1}^{1} \log\bigg(\frac{3-x}{3+x}\bigg) + \log\bigg(\frac{3+x}{3-x}\bigg) dx \\ 2I = \int_{-1}^{1} \log\bigg(\frac{3-x}{3+x} × \frac{3+x}{3-x}\bigg) dx \\ 2I=\int_{-1}^{1} \log(1) dx \\ 2I = \int_{-1}^{1} 0 \: dx \\ 2I = 0 \\ I = 0 \\ \text{therefore,} \\ \boxed{\int_{-1}^{1} \log\bigg(\frac{3-x}{3+x}\bigg) dx = 0}$