If $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}, x \neq 0$,then $\int_1^2 f\left(\frac{1}{x}\right) d x=$

Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x) = \begin{cases} x^2+3x+a, & x \leq 1 \\ bx+2, & x > 1 \end{cases}$ is differentiable on $\mathbb{R}$. Then,the value of $\int_{-2}^2 f(x) dx$ equals

Consider the equation $\int_1^e \frac{(\log_e x)^{1/2}}{x(a-(\log_e x)^{3/2})^2} dx = 1$,where $a \in (-\infty, 0) \cup (1, \infty)$. Which of the following statements is/are $TRUE$?

Evaluate the definite integral: $\int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x$

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

Let $f(x) = \max \{x+|x|, x-[x]\}$, where $[x]$ denotes the greatest integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) dx$ is