We start with completing the square in the denominator, and subsequently rearranging terms a little:
int \frac{2 x + 1}{\sqrt{x^2 + 10 x + 29}}\ d x =
= int \frac{2 x + 1}{\sqrt{( x + 5 )^2 + 4}}\ d x =
= int \frac{2 ( x + 5 ) - 9}{\sqrt{( x + 5 )^2 + 4}}\ d x.
By interpreting x + 5 as an inner function, with derivative 1, this is the same as
[ int \frac{2 t - 9}{\sqrt{t^2 + 4}}\ d t ]_{t = x + 5}.
Next, we substitute t \mapsto g ( u ) with
g ( u ) = 2 sinh u,
g' ( u ) = 2 cosh u, and
g^{-1} ( t ) = sinh^{-1} \frac{t}{2},
and thus end up with
[ int \frac{( 4 sinh u - 9 ) cosh u}{\sqrt{sinh^2 u + 1}}\ d u ]_{u = sinh^{-1} \frac{x + 5}{2}}.
Since \sqrt{sinh^2 u + 1} = cosh u, this is the same as
[ int 4 sinh u - 9 \ d u ]_{u = sinh^{-1} \frac{x + 5}{2}} =
= [ 4 cosh u - 9 u + C ]_{u = sinh^{-1} \frac{x + 5}{2}} =
= [ 4 \sqrt{sinh^2 u + 1} - 9 u + C ]_{u = sinh^{-1} \frac{x + 5}{2}} =
= 4 \sqrt{\frac{( x + 5 )^2}{4} + 1} - 9 sinh^{-1} \frac{x + 5}{2} + C =
= 2 \sqrt{( x + 5 )^2 + 4} - 9 sinh^{-1} \frac{x + 5}{2} + C =
=2 \sqrt{x^2 + 10 x + 29} - 9 sinh^{-1} \frac{x + 5}{2} + C.