How do I compute the improper integral $\int_0^{1/e}\frac{dx}{x(\log x)^{2}}$

Question

The given integral is $\int_0^{1/e}\frac{dx}{x(\log x)^{2}}$

ATTEMPT

I see that problem is at $0$, so I write the integral as

$$\lim_{t \to 0^{+}}\int_{0+t}^{1/e}\frac{dx}{x(\log x)^{2}}$$

Now I use the substitution, $\log(x)=u$. So now my integral becomes

$$\lim_{t \to 0^{+}}\int_{\log(t)}^{-1}\frac{du}{u^{2}}$$

Evaluating this gives me the answer $1$, but the textbook states that the answer us $8/3$. I would like to know where I went wrong

Thanks

Answer 1

Wolfram Alpha confirms that you are right. Your book probably has a misprint.