Is my integration of $\sqrt{4z^2 - 4z + 2}$ correct? I'm trying to
$$
\int \sqrt{4z^2 - 4z + 2}\ dz
$$
the integrand I first rewrite to (completing the square)
$$
4\left(z^2 - z +\frac{1}{4} - \frac{1}{4}\right) + 2 = 4\left(z - \frac{1}{2}\right)^2 + 1
$$
I apply $\int \sqrt{a^2 + x^2} \ dx = \frac{x}{2} \sqrt{a^2+x^2} + \frac{a^2}{2} \ln \left(x + \sqrt{a^2+x^2}\right)$.
So, before simplification, I get
$$
(z-1/2)\sqrt{4z^2-4z+2} + 1/2 \ln \left[ 2(z-1/2) + \sqrt{4z^2 - 4z +2} \right]
$$
According to my book, the solution is
$$
(z-1/2)\sqrt{z^2-z+1/2} + 1/4 \ln \left[ z - 1/2 + \sqrt{z^2 - z +1/2} \right]
$$
Is my solution equivalent to the book? I dont see how the book gets the `1/4' before the log.

You have made some little mistakes. I give you a hint to solve it more easier:
$$\int\sqrt{4z^2-4z+2}\space\text{d}z=$$
$$\int\sqrt{(2z-1)^2+1}\space\text{d}z=$$

Subsitute $u=2z-1$ and $\text{d}u=2\space\text{d}z$:

$$\frac{1}{2}\int\sqrt{u^2+1}\space\text{d}u$$

Subsitute $s=\arctan(u)$ and $\text{d}u=\sec^2(s)\space\text{d}s$:

$$\frac{1}{2}\int\sec^3(s)\space\text{d}s=$$
