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.

Let $x=\sqrt{7}\tan{u}\implies dx=\sqrt{7}\sec^2{u}du$
\begin{align}
I
=\int \frac{1}{(2x+1)\sqrt{x^2+7}}dx=\int \frac{\sec{u}}{(1+2\sqrt{7}\tan{u})}du\\
\end{align}
Let $t=\tan{\frac{u}{2}}\implies du=\frac{2}{1+t^2}dt$
\begin{align}
I
=\int\frac{\frac{1+t^2}{1-t^2}}{1+2\sqrt{7}\frac{2t}{1-t^2}}\frac{2}{1+t^2}dt=-2\int \frac{1}{t^2-4\sqrt{7}t-1}dt
\end{align}
Decompose the integrand into partial fractions and integrate.
\begin{align}
I=-2A\ln{(t-2\sqrt{7}-\sqrt{29})}-2B\ln{(t-2\sqrt{7}+\sqrt{29})}+c
\end{align}
where
\begin{align}
A={\rm Res}(f,2\sqrt{7}+\sqrt{29})=\frac{1}{2\sqrt{29}}, B={\rm Res}(f,2\sqrt{7}-\sqrt{29})=-\frac{1}{2\sqrt{29}}
\end{align}
Now express everything in terms of x.
