Solve the indefinite integral of 1/sqrt(x^2+2x+5).

Question

I solved it all out by completing the square and then trig sub and then drawing a triangle to go back to x variable and got:
ln |(sqrt(x^2+2x+5)/2 + (X+1)/2 ) |

but the book answer is:
ln |(sqrt(x^2+2x+5) + x + 1 ) |

So, how did they rid of the 1/2 or where did I go wrong?

Answer 1

1/√(x^2+2x+5) = 1/√((x+1)^2 + 3)

Now, ∫1/√(u^2+a^2) du = ln(x+√(u^2+a^2))

let u = x+2 and a^2=3, and you have a result of

ln(u+√(u^2+a^2)) = ln((x+1)+√((x+1)^2+3)
= ln(x+1+√(x^2+2x+5))

Answer 2

I don't understand what you mean?

Answer 3

This is how another person explained it to me:

dx/sqrt[(x+1)^2 +4 ]

let z = x+1
dx = dz

dz/sqrt[z^2 + 4]

= ln[z+sqrt(z^2+4)]

= ln[x+1 + sqrt(x^2+2x+1 + 4)]

because
integral dx/sqrt(x^2+p^2)
=ln[ x +sqrt(x^2+p^2)]

I still don't get how the 1/2 disappeared?