$\int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$ is equal to :

It is given that $\frac{d}{dt}(t \log t - t) = \log t$. Then, $\exp \left( \int_0^1 2x \log(1+x^2) dx \right) = $

$\int_0^a {{x^2}\sin {x^3}\,dx} $ equals

$\int_0^{\pi /4} \sec^7 \theta \sin^3 \theta \, d\theta = $

Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$,respectively with the positive $x$-axis. If $27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}$,where $\alpha$ and $\beta$ are integers,then the value of $\alpha+\beta$ equals:

The integral $\int_{\pi /6}^{\pi /3} {\sec ^{2/3} x \, \csc ^{4/3} x \, dx}$ is equal to