If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

If ${I_n} = \int_{ - n}^n {{{\tan }^2}\{x\}dx} $ then (where $\{.\}$ denotes the fractional part function and $n \in N$):

The number of continuous functions $f:[0,1] \rightarrow(-\infty, \infty)$ satisfying the condition $\int_0^1 (f(x))^2 dx = 2 \int_0^1 f(x) dx$ is

Let $\operatorname{Max} \limits _{0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\operatorname{Min} \limits _ {0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\beta$. If $\int\limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{e}\left(\frac{8}{15}\right)$,then $\alpha_{1}+\alpha_{2}$ is equal to

Let $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n x \, dx$. Then $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \dots$ are in:

Let $[t]$ denote the greatest integer less than or equal to $t.$ Then,the value of the integral $\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$ is equal to