$\int_0^1 tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ is equal to

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

The value of the definite integral $I = \int_0^2 x\sqrt{2-x} dx$ is:

If $I_n = \int_{0}^{\pi/4} \tan^n x dx$ then $I_{2024} + I_{2026}$ is equal to:

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

$\int_{3\pi/8}^{\pi/8} \frac{ \tan^{2025} x}{ \tan^{2025} x + \cot^{2025} x} dx$ is equal to

$\int \left(x + \frac{1}{x}\right)^2 dx$ equals :