Lời giải:
\(I=\int ^1_{-1}\frac{1}{(2-x)(2+x)}dx=\frac{1}{4}\int ^1_{-1}(\frac{1}{2-x}+\frac{1}{2+x})dx\)
\(=\frac{1}{4}\int ^1_{-1}\frac{1}{2-x}dx+\frac{1}{4}\int ^1_{-1}\frac{1}{2+x}dx\)
\(=\frac{-1}{4}\int ^1_{-1}\frac{1}{x-2}dx+\frac{1}{4}\int ^1_{-1}\frac{1}{x+2}dx\)
\(=(-\frac{1}{4}\ln |x-2|+\frac{1}{4}\ln |x+2|)^1_{-1}=\frac{1}{2}\ln 3\)