Problem: Find the UMVUE of $\frac1\theta$ for a random sample from the population distribution with density $$f(x;\theta)=\theta x^{\theta-1}$$ and show that its variance reaches the Cramér–Rao lower bound.
Progress: Knowing that the Cramér–Rao Lower Bound is $\frac{\gamma(\theta)^2}{nI(\theta)}$, where $\gamma(\theta)^2=\left(-\frac1{\theta^2}\right)^2=\frac1{\theta^4}$, I then compute the Fisher Information $I(\theta)$. This is where I get lost and become unsure about my work, mostly due to computation.
$$ \begin{align} I(\theta)&=\mathbb{E}\left(\frac{\partial\log{f(x;\theta)}}{\partial\theta}\right)^2\\ &=\int_0^1\left(\frac{\frac{\partial f(x;\theta)}{\partial \theta}}{f(x;\theta)}\right)^2f(x;\theta)\partial x \\ &=\int_0^1\frac{\left(\frac{\partial f(x;\theta)}{\partial \theta}\right)^2}{f(x;\theta)}\partial x \\ &=\int_0^1\left(\frac{1}{\theta}+\log{x})(x^{\theta-1}+(\theta^2-\theta)x^{\theta-2}\right)\partial x \\ &=\left[\log{x}\frac{x^\theta}{\theta}-\frac{x^{\theta-1}}{\theta-1}+\theta(\log{x})x^{\theta-1}\right]\biggr\rvert_0^1\\ &=\frac{-1}{\theta-1} \end{align} $$
I computed this and consistently got a negative answer, which doesn't make sense as plugging that into the CRLB formula yields a negative variance. I'd really appreciate if someone could quickly help me out on this!
Edit: I found out my mistake! It's a rather stupid one at the fourth equal sign: $$=\int_0^1\left(\frac{\partial\log{f(x;\theta)}}{\partial\theta}\right)\left(\frac{\partial f(x;\theta)}{\partial\theta}\right)$$ where I partially integrate the second term with respect to $x$ rather to to $\theta$.