<s>[INST] 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?
 [/INST] $$\int \frac{x-1}{9x^2-18x+17}dx=\dfrac{1}{18}\log(9x^{2}-18x+17)+Cst$$
indeed,


*

*Substitute $u=9x^2-18x+17$ and $du=(18x-18)dx$


\begin{align}
\int \frac{x-1}{9x^2-18x+17}dx&=\dfrac{1}{18}\int \dfrac{1}{u}du\\
&=\dfrac{1}{18}\log(u)+Cst
\end{align}


*

*Substitute back for $u=9x^2-18x+17$:
\begin{align}
\int \frac{x-1}{9x^2-18x+17}dx&=\dfrac{1}{18}\log(u)+Cst\\
&=\dfrac{1}{18}\log(9x^2-18x+17 )+Cst
\end{align}

</s>