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

If $g(x)$ is the inverse of $f(x)$ and $f(x)$ has domain $x \in [1, 5]$,where $f(1) = 2$ and $f(5) = 10$,then the value of $\int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy$ is equal to:

$\int_2^8 \frac{5^{\sqrt{10-x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x$ is equal to

$\int_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx = $

Let $f(x)$ be a function satisfying $f(x) + f(\pi - x) = \pi^2, \forall x \in R$. Then $\int_{0}^{\pi} f(x) \sin x \, dx$ is equal to $...........$.

If $\int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx = k \int\limits_0^\pi \ln(1 + \cos x) dx$,then the value of $k$ is: