Two apparently different evaluations of $\int \frac{x-1}{9x^2-18x+17}dx$ 
Evaluate the indefinite integral $$\int \frac{x-1}{9x^2-18x+17} \, dx .$$

This is an exercise from a book I'm studying. It gives the answer as: 
$$\ln(9x^2 -18x+17)^\frac{1}{18} +C .$$
This is an easy integral. You just notice that the numerator is the derivative of the denominator.
But I didn't notice exactly that at first, so I solved it in a slightly different way. I did:
$$\int \frac{x-1}{9x^2-18x+17}\, dx=\frac{1}{18}\int \frac{x-1}{\frac{x^2}{2}-x+\frac{17}{18}}\, dx$$ so $$z=\frac{x^2}{2}-x+\frac{17}{18}, \quad \frac{dz}{dx}=x-1 $$
I find the answer in the usual way as: $$\frac{1}{18} \int\frac{1}{z} dz= \frac{1}{18}\ln\left(\frac{x^2}{2}-x+\frac{17}{18}\right)+C , $$
which is (I believe) different from the answer the book gives, because the arguments of $ln$ are different. What is the problem here? Where am I wrong?

Despite appearances to the contrary, there's no problem here: The two expressions actually differ by a constant, but this equivalence is buried in a few special identities involving the logarithm function, namely $$\log (ab) = \log a + \log b$$ and $\log (a^b) = b \log a$ (both for appropriate $a, b$).
More specifically, note that
\begin{align}
\frac{1}{18} \log \left(\frac{x^2}{2} - x + \frac{17}{18}\right) + \color{#0000ff}{\frac{1}{18} \log 18}
&= \frac{1}{18}\log\left[\left(\frac{x^2}{2} - x + \frac{17}{18}\right) (18)\right] \\
&= \frac{1}{18} \log\left(9x^2 - x + \frac{17}{18}\right) \\
&= \log\left[\left(9x^2 - x + \frac{17}{18}\right)^{\frac{1}{18}}\right] .
\end{align}
By absorbing the constant $\color{#0000ff}{\frac{1}{18} \log 18}$ into $C$, we see that your general antiderivative and the one the text gives, regarded as families of functions (all equal up to addition of a constant), are actually the same.
