$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx =$

Let $f$ be a real-valued continuous function defined on the positive real axis such that $g(x) = \int_0^x t f(t) dt$. If $g(x^3) = x^6 + x^7$,then the value of $\sum_{r=1}^{15} f(r^3)$ is:

The approximate value of $\int_2^{10} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

The integral $\int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$ is equal to