Solving $\int\frac{\frac{1}{x}}{1-\frac{1}{x}}\;dx$ in the simplest way I am trying to solve a problem in my book:

$$\int\frac{\frac{1}{x}}{1-\frac{1}{x}}\;dx$$ 

I am trying to do this without using any techniques (eg. no u-sub, no parts...) just using what I know about basic integrals, by inspection. I tried to pull the denominator into the numerator to get some constant that I can pull in front of the integrand, but I have been unsuccessful. I am looking to see how I should simplify this and break it apart into smaller integrals that are easy to compute.
Thank You.
 A: The short answer has already been given; this one encompasses a few of the remarks and subtleties.
Note that:
$$\frac{\frac{1}{x}}{1-\frac{1}{x}}$$
is defined for $x \in \mathbb{R}\setminus \left\{ 0,1 \right\}$. For $x \ne 0$, rewrite:
$$\frac{\frac{1}{x}}{1-\frac{1}{x}} = \frac{\frac{1}{x}}{1-\frac{1}{x}}\frac{x}{x}=\frac{1}{x-1}$$
Now notice that:


*

*for $x>1$, you have:
$$\bigl( \ln(x-1) \bigr)' = \frac{1}{x-1}$$

*for $x<1$, you have:
$$\bigl( \ln(1-x) \bigr)' = \frac{-1}{1-x} = \frac{1}{x-1}$$


This is often abbreviated as:
$$\int \frac{1}{x-1} \,\mbox{d}x = \ln\left| x-1 \right| + C$$
but the following would be more accurate:
$$\int \frac{\frac{1}{x}}{1-\frac{1}{x}} \,\mbox{d}x =
\begin{cases}
\ln(x-1) + C_1 & \mbox{for } x > 1 \\
\ln(1-x) + C_2 & \mbox{for } 0 < x < 1 \\
\ln(1-x) + C_3 & \mbox{for } x < 0 \\
\end{cases}$$
In this way, you have the set of all possible anti-derivatives (since the constants need not be the same on the different intervals of the domain), which is what is usually meant with the "indefinite integral".
Of course for any definite integral on an interval $[a,b]$ not containing $x=0$ and $x=1$, you pick an appropriate anti-derivative.
A: Your function is
$$f (x)=\frac {1}{x-1} $$
the anti-derivative is
$$F (x)=\ln (|x-1|)+C. $$
$F $ is defined in an interval which doesn't contain neither $0$ nor $1$.
A: The given function can be simplified like this:



f(x) = 1/(x-1) 



The integral of this is 


F(x) = ln|x-1| + C.  
Where x is not defined for 0 or 1.


This is the easiest method for this integral.
