Find the value of $\int_0^1{4dx\over 4x^2-8x+3}$ 
Find the value of $\displaystyle\int_0^1{4dx\over 4x^2-8x+3}$

$$\begin{align*}\int_0^1{4dx\over 4x^2-8x+3}&=\int_0^1{dx\over (x-1)^2-(\frac 12)^2}
\\&=\int_0^1{dx\over (x)^2-(\frac 12)^2}
\\&=\int_{0}^{1/2}{dx\over (x)^2-(\frac 12)^2}+\int_{1/2}^{1}{dx\over (x)^2-(\frac 12)^2}
\\&=\bigg[\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg]_0^{1/2}+\bigg[\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg]_{1/2}^{1}
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg)-\ln(1)+\ln(\frac13)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x+\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x+\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\ln(1)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\ln(1)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{(1-x)-\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\ln(\frac13)
\end{align*}$$
Can someone please tell me where I made the mistake (wolframalpha says that the integral does not converge). Also how do we know the integral does not converge? Any help will be appreciated.
 A: Find the value of $\int_{0}^{1} \frac{4}{4x^2-8x+3} dx$
$$\int_{0}^{1} \frac{4}{4x^2-8x+3} dx = 4 \int_{0}^{1} \frac{1}{4x^2-8x+3} dx $$
Splitting the denominator we get,
$$4 \int_{0}^{1} \frac{1}{4x^2-8x+3} dx = 4 \int_{0}^{1} \frac{1}{(2x-3)(2x-1)} dx$$
Now using the method f partial fractions we get,
$$ 4 \int_{0}^{1} \frac{1}{(2x-3)(2x-1)} dx =  4 \int_{0}^{1} \frac{1}{2(2x-3)} -\frac{1}{2(2x-1)} dx = 4 \left[ \frac{1}{2} \int_{0}^{1} \frac{1}{2x-3} dx - \frac{1}{2} \int_{0}^{1} \frac{1}{2x-1} dx \right] = 2 \left[ \int_{0}^{1} \frac{1}{2x-3} dx -  \int_{0}^{1} \frac{1}{2x-1} dx \right]$$
And,
$$\int_{0}^{1} \frac{1}{2x-3} dx = \left[\frac{1}{2} \ln|2x-3|\right|_{0}^{1} = \frac{1}{2} \left( \ln|-1|- \ln|-3| \right)~~~~~~~~~ \textbf{(1)} $$
$$\int_{0}^{1} \frac{1}{2x-1} dx = \left[ \frac{1}{2} \ln|2x-1| |\right|_{0}^{1}  = \frac{1}{2} \left( \ln|1|- \ln|-1| \right)= \frac{1}{2} \ln|-1|~~~~~~~~~ \textbf{(2)}$$
Thus we have, $$2 \left[ \int_{0}^{1} \frac{1}{2x-3} dx -  \int_{0}^{1} \frac{1}{2x-1} dx \right] = 2 \left[ \frac{1}{2} \left( \ln|-1|-\ln|-3| \right) -\frac{1}{2} \ln|-1| \right] $$ $$ = -\ln|-3| = \ln \left( \frac{1}{3} \right)$$
You can use Cauchy principal value to verify that the integral diverges.
https://en.wikipedia.org/wiki/Cauchy_principal_value
