Two apparently different evaluations of $\int \frac{x-1}{9x^2-18x+17}dx$

Question

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?

Answer 1

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.

Answer 2

$$\int \frac{x-1}{9x^2-18x+17}dx=\dfrac{1}{18}\log(9x^{2}-18x+17)+Cst$$

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}