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,
$$\Rightarrow \bbox[5px,border:2px solid red]{\begin{align}I\;=\;&2x-\sqrt{4x^2-x+1}+4\ln(\sqrt{4x^2-x+1}-2x+4)\\
&-\frac{1}{4}\ln(4\sqrt{4x^2-x+1}-8x+1)+C\,.
\end{align}}$$
Hint
$$I=\int{\frac{dx}{(x^2+4)^3}}$$
$$x=2\tan(t) \implies dx=2 \sec ^2(t)\implies I=\frac 1{32}\int \cos^4(t)\,dt$$
Using twice the double angle formulae,
$$\cos^4(t)=\frac{3}{8}+\frac{1}{2} \cos (2 t)+\frac{1}{8} \cos (4 t)$$
Best Answer
A nice trick is to write $\frac{x}{(1+x)^2}$ as $\frac{(x+1)-1}{(x+1)^2}=\frac{1}{(x+1)}-\frac{1}{(x+1)^2}$, then apply integration by parts:
$$ \int \frac{e^x}{(1+x)^2}\,dx = -\frac{e^x}{1+x}+\int \frac{e^x}{(1+x)}\,dx.$$