$$ \int \frac1x = \log_e |x|+C$$
Why is modulus sign needed. If this is because the domain of logarithmic function is $(0,\infty)$ Then why don't we mention the limitations of the domains of other functions like tan, cot, sec, cosec etc. while solving integrals?
The derivative of $\ln(x)$ is $\frac{1}{x}$ for $x>0$. The derivative of $\ln(-x)$ is $\frac{1}{-x}(-1)=\frac{1}{x}$ for $x<0$.
So we can combine these cases and say that the derivative of $\ln|x|$ is $\frac{1}{x}$ for $x\ne 0$, since $|x|$ is either $x$ or $-x$.
Thus, the antiderivative of $\frac{1}{x}$ is $\ln|x|+C$. If you don't put the absolute value, it's not wrong in all cases. You'd be fine calculating $\int_1^2\frac{1}{x}dx$, but you'll have problems calculating $\int_{-2}^{-1}\frac{1}{x}dx$. Putting the absolute values just provides a more complete picture of your antiderivative--one that is valid on a larger domain.
Edit: Regarding the other functions, they are defined everywhere except for a discrete set of points (rather than being defined for positive and not negative values). I don't see how the situation is similar to that of the log function. The antiderivative of $\sec^2(x)$ is $\tan(x)+C$. If you argue that it is $\tan|x|+C$, then you'd have to see what happens if $x<0$. If this is true, then $|x|=-x$, so the derivative of $\tan(-x)$ is $-\sec^2(-x)=-\sec^2(x)$, since sec is even, and this is not the same as $\sec^2(x)$. So the idea just doesn't extend.
Another note: the function $\frac{1}{x}$ has domain $(-\infty,0)\cup(0,\infty)$, while the function $\ln(x)$ has domain $(0,\infty)$. So in a sense, the equation $\int \frac{1}{x}dx=\ln(x)+C$ would be a bit lacking because the domains don't match. Looking at $\int\sec^2(x)dx=\tan(x)+C$, however, both $\sec(x)$ and $\tan(x)$ have the same domain already, so there really is no mismatch we can cleverly make use of like we did before.
The notation $\int dx/x$ is supposed to mean "the general antiderivative of $1/x$ on its domain," which is $\mathbf R - \{0\} = (-\infty,0) \cup (0,\infty)$, and the right answer should have separate additive constants on each of the intervals: $$ \int \frac{1}{x}\,dx = \begin{cases} \ln x + C_1, & \text{ if } x > 0 \\ \ln|x| + C_2, & \text{ if } x < 0 \end{cases} = \begin{cases} \ln|x| + C_1, & \text{ if } x > 0 \\ \ln|x| + C_2, & \text{ if } x < 0, \end{cases} $$ where $C_1$ and $C_2$ are constants that need not be equal. However, in practice the way that this antiderivative formula is used in standard single-variable calculus courses is to work with an antiderivative of $1/x$ on an interval inside its domain, which makes this interval a subinterval of $(0,\infty)$ or a subinterval of $(-\infty,0)$, and for this purpose only a single additive constant is needed. Thus in practice no error will be made by using the incomplete formula $$ \int \frac{1}{x}\,dx = \ln|x| + C $$ on $\mathbf R - \{0\}$, where $C$ is a constant. The constant is totally irrelevant to calculate $\int_a^b dx/x$ with $(a,b)$ in $(-\infty,0)$ or $(0,\infty)$, since a single antiderivative is needed on that interval, and $\ln|x|$ can serve this role.
In a similar way, since the domain of $\tan x$ in $\mathbf R$ is a union of infinitely many disjoint open intervals $I_k = (-\pi/2 + \pi k,\pi/2 + \pi k)$ where $k$ runs through the integers, the correct general antiderivative formula for $\tan x$ in a single-variable calculus course should be $$ \int \tan x \,dx = -\ln|\cos x| + C_k \text{ for } k \in \mathbf Z, $$ where the $C_k$'s are separate constants on each of the intervals $I_k$. These constants need not all be the same. If the only purpose to which a student in a course ever puts an antiderivative formula is to compute $\int_a^b \tan x\,dx$ where $\tan x$ is defined on $(a,b)$ (no asymptote in $(a,b)$), then the additive constant ambiguity is irrelevant and the simpler formula $-\ln|\cos x|$ can be used, or equivalently $\ln|\sec x|$.
The derivative of $\log |x|$ is $\dfrac 1x$ at any point $x \not= 0$. The fundamental theorem of calculus tells you that $$\int_a^b \frac 1t \, dt = \log |b| - \log |a|$$ for any interval $[a,b] \subset (-\infty,0) \cup (0,\infty)$.
