Finding a definite integral of $(\frac{x}{2x-1})$ Once again I'm doing some preparatory work for a course and have hit a stumbling block and would appreciate some pointers.  I think I'm most of the way there...
So the question is calculate the following:
$$\int_0^1 \frac{x}{2x-1}$$
Which to me is definite integrals.  My first thought was to express as partial fractions in order to make it easier to apply rules, but as the numerator must be a lower degree than the denominator, I must do a long division, which I did as follows:
$$x/(2x-1) $$
Which gives , simplified:
$\frac{1}{2}. \frac{1/2}{2x-1}$ working out as:
$$ (\frac{1}{2})(\frac{1}{2(2x-1)}) $$
Now I can apply the fact that $\frac{1}{ax+b} = \frac{1}{a}\ln(ax+b) + C$, AND integrate individual terms, ADN factor out common items meaning:
$$ (\frac{1}{2})(\frac{1}{2(2x-1)}) = \frac{1}{2}\int\frac{1}{2x-1}dx + \frac{1}{2}\int1dx$$
Which works out on paper to:
$$\frac{1}{2}x + \frac{1}{4}\ln(2x-1) + C $$
But then if you attempt to substitute in your upper and lower limits, you are trying to take the natural log of zero, which is a maths error as far as I can see? 
I saw a similar question here: Integral of $x/(2x-1)$ But thats not dealing with limits.  Is this a trick question or something?  Regardless any help would be greatly appreciated.
 A: The integrand is $$\frac12\left(1+\frac{1/2}{x-1/2}\right).$$ Thus the integral is $$\frac12\left(x+\frac12\log|x-1/2|\right)+K.$$
This exists both at $0$ and $1.$
A: The problem is the $\frac 12$ is a pole in this integral.
First note that it should be $\frac 12x+\frac 14\ln|2x-1|+C$ with a different constant on any interval not containing the pole.
So you have to study $\displaystyle I(a,b)=\int_0^{\frac 12-a}\frac x{2x-1}\mathop{dx}+\int_{\frac 12+b}^1\frac x{2x-1}\mathop{dx}$
If $\lim_{(a,b)\to(0^+,0^+)}I(a,b)$ exists then your integral is well defined.
This is not the case here since you'll get some $\ln(a)-\ln(b)$ which diverges.
You could also examine if $\lim_{a\to 0^+}I(a,a)$ exists, this "symmetrical" calculation is called the Cauchy principal value.
https://fr.wikipedia.org/wiki/Valeur_principale_de_Cauchy
This time, the logarithms cancel each other and we get $\frac 12$ as CPV.
A: Let $f(x)=\dfrac{x}{2x-1}.$  The function is undefined at $x=\frac12$, so we must rewrite our integral as
$$\int_0^1\frac{x}{2x-1}\ dx=\lim_{t\to\frac12^-}\int_0^t\frac{x}{2x-1}\ dx+\lim_{s\to\frac12^+}\int_s^1\frac{x}{2x-1}\ dx.$$
If this is for anything before Calc 2, you can simply say that this integral diverges.   But the Cauchy principal value for this integral will be $\frac12$.  Here is a video that might explain it more clearly.
