Evaluate the following integrals: Evaluate: . Hence, evaluate: .

Evaluating $\int \ln x dx$ using integration by parts: Let $u = \ln x$ and $dv = 1 dx \Rightarrow du = \frac{1}{x}dx$ and $v = x$. $\int \ln x dx = x\ln x - \int x\left(\frac{1}{x}\right)dx = x\ln x - \int 1 dx = x\ln x - x + C = x(\ln x - 1) + C$.

To evaluate $I = \int \frac{\ln[\ln(\frac{1+x}{1-x})]}{1-x^2} dx$: Let $t = \ln\left(\frac{1+x}{1-x}\right)$. Differentiating with respect to $x$: $\frac{dt}{dx} = \frac{1}{\frac{1+x}{1-x}} \cdot \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1-x}{1+x} \cdot \frac{2}{(1-x)^2} = \frac{2}{1-x^2}$. Thus, $\frac{dx}{1-x^2} = \frac{dt}{2}$. Substituting into the integral: $I = \int \ln t \cdot \frac{dt}{2} = \frac{1}{2} \int \ln t dt$. Using the result from part (a), $\int \ln t dt = t(\ln t - 1) + C$: $I = \frac{1}{2} t(\ln t - 1) + C = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \left[\ln\left(\ln\left(\frac{1+x}{1-x}\right)\right) - 1\right] + C$.