The value of the integral $\int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} dx$ is

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$.

Read the following mathematical statements carefully:
$I.$ $A$ differentiable function $f$ with maximum at $x = c$ $\implies f''(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$,$\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx}$.
$IV.$ If $f(x)$ has a maxima at $x = c$,then $f$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \to 0$ for $h > 0$. Now indicate the correct alternative.

$\int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} \, dx = $

$A$ function $f(x)$ satisfies $f(x) = f(\frac{c}{x})$ for some real number $c$ $(c > 1)$ and $\forall\, x > 0$. If $\int_{1}^{\sqrt{c}} \frac{f(x)}{x} dx = 3$,then the value of $\int_{1}^{c} \frac{f(x)}{x} dx$ is

For $x > 0,$ if $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt,$ then $f(e) + f\left(\frac{1}{e}\right)$ is equal to: