How to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $ Im trying to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $.
I've tried integration by parts, and any resonable substitution that came to my mind. Neither seem to work. I tried to use integral online calculator, the result was too complicated, and I'm pretty sure there is a resonable way to solve it, as we had this question in an exam.
Thanks in advance.
 A: Take,$$t=\sqrt{4x^2-x+1}-2x$$
Then,$$\sqrt{4x^2-x+1}=2x+t \;\;\Longrightarrow\;\;4x^2-x+1=(2x+t)^2 \;\;\Longrightarrow\;\; x=\frac{1-t^2}{4t+1}\,.$$
And $$dx = -\frac{4t^2+2t+4}{(4t+1)^2}dt\,.$$
So,
$$I=\int\frac{2x-\sqrt{4x^2-x+1}}{x-1}dx=\int\frac{t}{(1-t^2)/(4t+1)-1}\cdot\frac{4t^2+2t+4}{(4t+1)^2}dt\,,$$
$$\begin{align} & =-\int\frac{4t^2+2t+4}{(4t+1)(t+4)}dt\\
&=-\int\frac{4t^2+2t+4}{4t^2+17t+4}dt\\
&=-\int\left(1-\frac{15t}{4t^2+17t+4}\right)dt\\
&=-t+\int\frac{15t}{(4t+1)(t+4)}dt\,.
\end{align}$$
Splitting into partial fractions,
$$\frac{15t}{(4t+1)(t+4)} = \frac{4}{t+4}-\frac{1}{4t+1}\,,$$
We get,
$$I=-t+\int\frac{4}{t+4}dt-\int\frac{1}{4t+1}dt=-t+4\ln(t+4)-\frac{1}{4}\ln(4t+1)+C\,,$$
so, therefore,
$$\Rightarrow \bbox[5px,border:2px solid red]{I=-t+4\ln(t+4)-\frac{1}{4}\ln(4t+1)+C\,,}$$
On rewriting we get,
$$\Rightarrow \bbox[5px,border:2px solid red]{\begin{align}I\;=\;&2x-\sqrt{4x^2-x+1}+4\ln(\sqrt{4x^2-x+1}-2x+4)\\
&-\frac{1}{4}\ln(4\sqrt{4x^2-x+1}-8x+1)+C\,.
\end{align}}$$
