Integration of sum of infinite geometric series. It is given in CRM book, by MAA (A Radical Approach to Real Analysis By David M. Bressoud) in exercise of Q. 2.2.9  

It is tempting to integrate each side of eqn. (2.11) wrt $x$ and to assert that

The eqn. (2.11) is for sum of infinite geometric series as given by:
 $1+x+x^2+x^3+\cdots=\frac1{1-x}$, if $|x|\lt 1$.

$x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdot=-\ln(1-x).$ (2.13)
Following Cauchy's advice (on pg. 19), we know we need to be careful, but now we run into trouble.

This means that need take a finite lower bound on the number of terms ($n$). E.g., the eqn. (2.10) is for sum of geometric series, for a lower bound $n$ as given by: 
$1+x+x^2+x^3+\cdots+x^{n-1}=\frac1{1-x} - \frac {x^n}{1-x}$, if $|x|\lt 1$.

What happens when we try to integrate $\frac{x^n}{(1-x)}$?

Doubt 1: Not clear about it, as need to find integration using usual methods. But, if take it as a stand-alone problem, then how to pursue/ attack it.

Fortunately, we do not have to find the exact value of the difference between $ x+\frac{x^2}2+\frac{x^3}3+\cdots+\frac{x^n}n$ and $-\ln(1-x)$. All we have to show is that we can make this difference as small as we wish by taking enough terms. We can do this by bounding the integral of $\frac{x^n}{(1-x)}$.

Need find bounds $a,b$ s.t. form limits for sum of enough terms, denoted by the difference $|b-a|$ 
$\int_a^b \frac{x^n}{(1-x)}dx$

We use the fact that if
$|f(x)|\lt g(x) $ for all x, then $|\int_a^b f(x)dx|\lt \int_a^b g(x)dx$.
 If $0\lt x \lt 1$, then we can find a number $a$ so that $0\lt x\lt a\lt 1$ and 
$|\frac{x^n}{1-x}| \lt x^n(1-a)^{-1}$

Doubt 2: Unable to see the significance of absolute value of limits of integrand obtained for $f(x)$. 
The inequality $|f(x)|\lt g(x)$ shows that for function $f(x)$ the negative values taken are also summed up. 
Is this a well-known result, if so what is its name?
Also, what is the significance of it here.

Integrate this bounding function w.r.t. $x$, & show that if $0\lt x \lt 1$, then the partial sums of the series in eqn. (2.13) approach the target value of $-\ln(1-x)$ as $x$ increases.

Doubt 3:  Unable to pursue this computation. Need some clue.

Explain what happens for $-1\lt x\lt 0$. Justify your answer.

Doubt 4: Unable to find the significance of such bounds.
 A: We could have used the definition of incomplete Beta function
$$B(z; a,b) = \int_0^z u^{a-1}(1-u)^{b-1}\, du$$
to express
$$\int_0^{z}\frac{u^n}{1-u}\, du = \int_0^z u^{(n+1)-1}(1-u)^{-1}\, du = B(z; n+1, 0).$$
but if you don't understand the property of incomplete Beta function, it doesn't give you good control. 
Also, for $|x| < 1$,
$$\sum_{i=0}^{n-1}x^i = \frac1{1-x}- \frac{x^n}{1-x}$$
$$\int \frac{x^n}{1-x}\, dx = -\ln (1-x) - \sum_{i=1}^{n}\frac{x^i}{i}+C$$
which is exactly the error term. 
The message that the book is trying to send is you don't have to work with this quantity directly. 
$$\int_{0}^x \frac{t^n}{1-t}\, dt = \int_0^x\left(\frac{1}{1-t}-\sum_{i=0}^{n-1}t^i \right)\, dt=-\ln (1-x) - \sum_{i=1}^n \frac{x^i}{i}$$
The error term is obtained by taking the absolute value,
$$\left| \int_{0}^x \frac{t^n}{1-t}\, dt\right|.$$
So our goal is to show that as $n$ is big enough, the error term shrink to zero. 
Fixing a value of $x \in (0,1)$.
$$\left| \int_{0}^x \frac{t^n}{1-t}\, dt\right| \le \int_0^x \frac{t^n}{1-x} \, dt = \frac{x^{n+1}}{(n+1)(1-x)} \to 0$$
Fixing a value of $x \in (-1, 0)$, 
$$\left| \int_{0}^x \frac{t^n}{1-t}\, dt\right| \le \int_x^0 |t|^n\, dt = \frac{|x|^{n+1}}{n+1}\to 0$$
