(a)

Evaluate : \[\begin{gathered} \int_{-2}^{2} \sqrt{\frac{2-x}{2+x}} \mathrm{dx} \\ \text { OR } \end{gathered} \]

(b)

Find : \[\int \frac{1}{x\left[(\log x)^{2}-3 \log x-4\right]} d x \]

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