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.