(a) Evaluate: ∫_-2^2 √((2-x)/(2+x)) dx; OR (b) Find: ∫ 1/(x[( log x)^2-3 log x-4]) d x

$$\begin{aligned} & \text { (a) } \int_{-2}^{2} \sqrt{\frac{2-x}{2+x}} d x \\ & =\int_{-2}^{2} \frac{2-x}{\sqrt{4-x^{2}}} d x \\ & =\int_{-2}^{2} \frac{2}{\sqrt{4-x^{2}}} d x-\int_{-2}^{2} \frac{x}{\sqrt{4-x^{2}}} d x \end{aligned} $$ $\displaystyle =2 \int_{0}^{2} \frac{2}{\sqrt{4-x^{2}}} d x-0\left[\frac{2}{\sqrt{4-x^{2}}}\right.$ is even, $\displaystyle \frac{x}{\sqrt{4-x^{2}}}$ is odd $\displaystyle ]$ $$=4 \int_{0}^{2} \frac{1}{\sqrt{4-x^{2}}} d x $$ $\displaystyle =2 \pi$ Let $\displaystyle \log x=t \Rightarrow \frac{1}{x} d x=d t$ The given integral becomes $\displaystyle =\int \frac{1}{t^{2}-3 t-4} d t$ $$=\int \frac{1}{\left(t-\dfrac{3}{2}\right)^{2}-\left(\dfrac{5}{2}\right)^{2}} d x $$