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#.
