#int(2x+1)/sqrt(x^2+10x+29)dx#?

1 Answer
May 11, 2018

#int \frac{2 x + 1}{\sqrt{x^2 + 10 x + 29}}\ d x =#
#=2 \sqrt{x^2 + 10 x + 29} - 9 sinh^{-1} \frac{x + 5}{2} + C#

Explanation:

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#.