$\int_{\frac{1}{3}}^{1} \frac{(x - x^3)^{\frac{1}{3}}}{x^4} dx =$

$\displaystyle\int e^x(\sec x + \log_e(\sec x + \tan x))dx$ is equal to: (where C is an arbitrary constant)

$\int \frac{\sqrt{1+x^2}}{x^4} dx$ is equal to :

$\int_1^{\sqrt{3}} \frac{1}{1+x^2} dx$ is equal to:

If $\int \sqrt\frac{1-x}{{1+x}} dx = a\sqrt{1-x^2} + \beta \sin^{-1}x + C$, Where C is an arbitrary constant, then which of the following are TRUE?

(A) $\alpha = 1$

(B) $\alpha = -1$

(C) $\beta = 1$

(D) $\beta = -1$

Choose the correct answer from the options given below:

If $\displaystyle\int_0^{40} \dfrac{dx}{2x+1} = \log_e k$, then the value of $k$ 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