Need help in indefinite integral elimination I'm trying to get a easy and short answer for that indefinite integral:
$$\int\frac{\sqrt{x^2+x+1}}{(x+1)^2}\,dx$$
But everything I find is very big and difficult to understand (too much so it seems like a wrong result). Is there any way to solve it in a more beautiful way?

If we let $$p=2\frac{\sqrt{x^2+x+1}}{x+1},q=\frac{x-1}{x+1}
$$So $p^2=3+q^2$ with $p\text{d}p=q\text{d}q$.And
\begin{aligned}\int{\frac{x\mathrm{d}x}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}}&=\frac{1}{2}\int{\frac{q+1}{p}\mathrm{d}q}
\\
&=\frac{1}{2}\int{\mathrm{d}p}+\frac{1}{2}\int{\frac{\mathrm{d}q}{p}}
\\
&=\frac{1}{2}p+\frac{1}{2}\int{\frac{\mathrm{d}q+\mathrm{d}p}{p+q}}
\\
&=\frac{1}{2}p+\frac{1}{2}\ln \left( p+q \right) 
\end{aligned}
At this time,indefinite integral can be easily done:
\begin{align}&\int{\frac{\sqrt{x^2+x+1}}{\left( 1+x \right) ^2}\mathrm{d}x}=\int{\frac{x^2+x+1}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}\mathrm{d}x}
\\
&=\int{\frac{\mathrm{d}x}{\sqrt{x^2+x+1}}}-\int{\frac{x\mathrm{d}x}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}}
\\
&=\ln \left( x+\frac{1}{2}+\sqrt{x^2+x+1} \right) -\frac{\sqrt{x^2+x+1}}{x+1}-\frac{1}{2}\ln \left( \frac{x-1+2\sqrt{x^2+x+1}}{x+1} \right) 
\end{align}
