$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{tanx}}$ is equal to

The integral I = $\int e^x\left(\frac{x - 1}{3x^2}\right) dx$ is equal to

$\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx =$

$\int_{\pi/6}^{\pi/3} \frac{tan x}{tan x + cot x} dx$ 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:

$\int \frac{1}{(x + 1)(x + 2)} dx$ is equal to

If $\int \frac{1}{\tan x + \cot x}dx = k\cos 2x + C$, where C is an arbitrary constant, then the value of $k$ is