$\int e^x \left(\frac{x-1}{2x^2}\right) dx$ is equal to ( where $C$ is an arbitrary constant)

If $2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1$, then $\int f(x) dx$ is: (Here C is an arbitrary constant)

If $f(x)$ and $g(x)$ are continuous functions in [0, a] such that $f(x) = f(a - x)$ and $g(x) + g(a - x) = a$ then $\int_{0}^{a} f(x)g(x)dx =$

The value of the definite integral $\int_0^1 e^x \frac{(1-x)^2}{(1+x^2)^2}dx$ is:

If $\int (x + \sqrt{x^2 - 1})^2 \, dx = \alpha \cdot x + \beta x^3 + \gamma (x^2 - 1)^{\frac{3}{2}} + C$, where $C$ is arbitrary constant, then the value of $3(\alpha + \beta + \gamma)$ is

$\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx$ is equal to

$\int \frac{x^9}{(4x^2+1)^6}\,dx$ is equal to:

[where c is an arbitrary constant]