Calculating Arc Length of a Curve: A Calculus II Problem

[LBK]

Homework Statement

Find the exact length of the curve: y= 1/4 x²-1/2 ln(x) where 1<=x<=2

Homework Equations

Using the Length formula (Leibniz) given in my book, L=Int[a,b] sqrt(1+(dy/dx)²)
I found derivative of f to be (x²-1)/2x does that look correct?

The Attempt at a Solution

I found f' to be (x²-1)/2x does that look correct? I'm thinking I must be off on something because I get a wrong answer. Either that or my integration. I am trying to use tables in the book for that, and not sure about that either. I am trying to sub u for that (dy/dx) part when integrating, for which I think I can use the form:
int sqrt(u²+a²)=1/2 (u*sqrt(u²+a²)+a²ln|u+ sqrt(u²+a²)
When I put my value back in for u, it turns into a big mess. When I plug in my x values I got 15/32 +ln 2. This isn't right, the book has 3/4 + 1/2 ln 2 if that helps. Thank you.

 

[pasmith]

Write $y'$ in the simpler form $\frac12(x - \frac{1}{x})$. Take a closer look at $1 + y'^2$, and consider the identity $(a + b)^2 = (a - b)^2 + 4ab$.

 

[Mark44]

Like many textbook problems on arc length, this one is cooked up so that the square root simplifies to something nice. As it turns out, 1 + y'² is a perfect square, so the final integral doesn't involve a radical and is pretty simple. I get the same answer as the textbook.

 

[LBK]

ah, I get it now. Thanks, you guys rock. I'll definitely be back lol.