Asked by Dave
                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?
            
        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?
Answers
                    Answered by
            Steve
            
    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))
    
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))
                    Answered by
            Dave
            
    I don't understand what you mean?
    
                    Answered by
            Dave
            
    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?
    
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?
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