If $\int \frac{(\sqrt{x})^5}{(\sqrt{x})^7+x^6} dx = a\log\left|\frac{x^k}{1+x^k}\right| + C$; Where C is an arbitrary constant, then the value of $a.k$ is:

$\int_0^2 (|x| + |x - 2|) dx =$

$\displaystyle\int \dfrac{1+\cos\theta}{1-\cos\theta}\,d\theta =$

$\int \dfrac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx$, $x>0$ and $x\ne 1$ is equal to (consider $\log x = \log_e x$)

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:

The value of the integral $\int_{-6}^{0} |x+3|\,dx$ is

Value of $\int_{1}^{4} \frac{\log_e x}{x^2}\, dx$ is