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

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

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

For $x \in \mathbb{R} - \{-1,0,1\}$, $\int \frac{1}{x - x^5}dx$ is equal to

Which of the following is/are correct?

A. $\int e^x\left[\frac{1}{(x-2)}-\frac{1}{(x-2)^2}\right]dx=\frac{e^x}{(x-2)}+C$, where C is an arbitrary constant.

B. $\int e^x\left[\frac{1}{(x-2)^3}-\frac{3}{(x-2)^4}\right]dx=\frac{e^x}{(x-2)^2}+C$, where C is an arbitrary constant.

C. $\int e^x\left[\frac{x-4}{(x-2)^3}\right]dx=\frac{e^x}{(x-2)^2}+C$, where C is an arbitrary constant.

D. $\int e^x\left[\frac{1}{x+2}-\frac{3}{(x+2)^2}\right]dx=\frac{e^x}{(x+2)^2}+C$, where C is an arbitrary constant.

Choose the correct answer from the options given below:

$\int \sqrt{1 + \frac{x^2}{9}} dx$ is equal to

(Where C is an arbitrary constant)

If $\dfrac{d}{dx}[f(x)] = 5x^4 - \dfrac{2}{x^3}$ such that $f\left(-\dfrac{1}{2}\right) = 0$, then $f(x)$ is equal to