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