Why does the antiderivative of $\frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$)

Why does the antiderivative of $\frac{1}{x}$ have to be $0$ at precisely $x=1$?

Are you familiar with the fundamental theorem of calculus? In your case it can be stated as

$$\int_y^x \frac{1}{t}dt=\ln(x)-\ln(y)$$

Because $\ln(t)$ is an antiderivative of $\frac{1}{t}$. Now for $x=1$ we have $\ln(x)=0$ and thus

$$\int_1^x \frac{1}{t}dt=\ln(x)$$

So the right side is very simple. That's why the choice of $1$ is so convenient.

So the question should not be "why antiderivative is $0$ at $1$" but rather "why we chose $\ln(x)$ to be the very special antiderivative of $\frac{1}{x}$". And the reason is stated above. Simplicity.

EDIT: So now I understand that actually

$$\ln(x):=\int_1^x \frac{1}{t}dt$$

is the definition of logarithm in your book. So now the question "why we pick $1$?" makes more sense. The reason is not obvious even though it's simple: because it works. In the sense that this definition of $\ln(x)$ has some good properties, in particular it's a homomorphism

$$\ln(xy)=\ln(x)+\ln(y)$$

Also it's the inverse of the classical exponential map:

$$e^x:=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

These properties won't hold if you use any other number then $1$ in the integral.

I understand the confusion now. And IMO it is not good (from educational point of view) to define the logarithm like that.

Side note: starting from $0$ is not good since $\int_{0}^x\frac{1}{t}dt=\infty$ for any $x>0$. This corresponds to the fact that the tangent line to $\ln(x)$ at $0$ is vertical.

From your comments it seems like the part of your question not yet answered is "why is $\log(1)$ defined to be $0$? The answer here lies in the purpose of the logarithm.

Logarithms convert multiplication into addition: $\log(xy) = \log(x)+\log(y)$, and they do that because it's easier for humans to add things up than it is to multiply them together. In order for this to work the logarithm has to map the multiplicative identity to the additive identity: i.e. $log(1) = 0$. If it didn't then we'd get $\log(x) = \log(1\cdot x) = \log(1) + \log(x) \not= \log(x)$ which would be a problem.

Logarithms are often described (see wikipedia for example) as inverting the exponentiation operator, which maps addition into multiplication.

If you have a look at this page of epistemology (sorry it is in French butI'll try to summarize the main points)

https://fr.wikipedia.org/wiki/Histoire_des_logarithmes_et_des_exponentielles

The article says that at the beginning $\log(1)$ was not zero, but instead is was $\log(10^7)=0$

Around $1615-1616$ Briggs and Neper had some discussion about the pros and cons of their own choices about the new tool (the logarithms) and the point was that multiplying quantities by various powers of $10$ to fit such or such pre-calculated table was not pleasant, and it seems they agreed on setting $\log(1)=0$.

An additional advantage was the fundamental relation $\mathbf{\log(a\times b)=\log(a)+\log(b)}$

(for which $\log(1)=2\log(1)$ forces $\log(1)=0$)

This break through allowing to manipulate products likes sums via tables of anti-logarithms.$^*$

$(*)$ exponential was developed much later in history, with Euler and the value of $e$, standing for $\bar euler$ I presume...