$\int_{-1/2}^{1/2} \log \left(\frac{1+x}{1-x}\right) dx=$

Evaluate: $\int_{-a}^a f(x) dx - \int_0^a f(-x) dx$

If $[a]$ denotes the greatest integer which is less than or equal to $a$, then the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\sin x \cos x] dx$ is

If $f(x) = f(2 - x)$,then $\int_{0.5}^{1.5} xf(x) dx$ equals

$\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$, then $f(x) =$

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$,$x \in R - \{-1\}$,$n \in N$,$n > 2$. If $f^n(x) = (f \circ f \circ f \dots \text{upto } n \text{ times})(x)$,then $\lim_{n \to \infty} \int_0^1 x^{n-2} (f^n(x)) dx$ is equal to $...............$.