Evaluate $\int \frac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x$. How to do this indefinite integral (anti-derivative)?
$$I=\displaystyle\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x.$$
I tried doing some substitutions ($x^2+7=t^2$, $2x+1=t$, etc.) but it didn't work out.

Using Euler substitution by setting $t-x=\sqrt{x^2+7}$, we will obtain 
$x=\dfrac{t^2-7}{2t}$ and $dx=\dfrac{t^2+7}{2t^2}\ dt$, then the integral turns out to be
\begin{align}
\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\ dx&=\int\frac{1}{t^2+t-7}\ dt\\
&=\int\frac{1}{\left(t+\dfrac{\sqrt{29}+1}{2}\right)\left(t-\dfrac{\sqrt{29}-1}{2}\right)}\ dt\\
&=-\int\left[\frac{2}{\sqrt{29}(2t+\sqrt{29}+1)}+\frac{2}{\sqrt{29}(-2t+\sqrt{29}-1)}\right]\ dt.
\end{align}
The rest can be solved by using substitution $u=2t+\sqrt{29}+1$ and $v=-2t+\sqrt{29}-1$.
