$\int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ Comparison Test? So I am having a bit of trouble with this integral, and I would like some verification to see if I am on the right track.

Q: Determine if $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ converges or diverges.

So I know that we need to find a function that is comparable to the original function. So I chose $\displaystyle \dfrac{1}{x\ln(x)}$ because
\begin{equation*}
\dfrac{1}{x\ln(x + 1)} \leq \dfrac{1}{x\ln(x)}
\end{equation*}
for $x \geq 1$.
So we take the integral: $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x)} \ dx$. Let $u = \ln(x)$, then $du = \dfrac{1}{x} \ dx$.
\begin{align*}
\int_{1}^{\infty} \dfrac{1}{x\ln(x)} \ dx &= \int_{x = 1}^{x = \infty} \dfrac{1}{u} \ du \\
&=\lim_{t \to \infty}\left[\ln|u|\right]_{x = 1}^{x = t} \\
&= \lim_{t \to \infty} \left[\ln|\ln(x)|\right]_{1}^{t} \\
&= \lim_{t \to \infty} \ln|\ln(t)| - \lim_{t \to \infty} \ln|\ln(1)| \\
&= \infty - DNE
\end{align*}
The last part is where I am having a bit of trouble. I know that the integral diverges with the infinity, which means that $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ also diverges. But if $\ln|\ln(1)|$ does not exist, does that mean this integral still diverges? I am not sure how to explain this part well. Would appreciate some tips.


OP wrote:  I know that the integral diverges with the infinity, which means that $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ also diverges.

You can't figure this out of comparison test. The comparison test says that:
When for all $x$ we have $0 ≤ f(x) ≤ g(x)$, if $\displaystyle \int f(x) \ dx$ diverges, than $\displaystyle \int g(x) \ dx$ diverges, but not the opposite!
I think this is what you're looking for
$\begin{equation*}
\dfrac{1}{(x+1)\ln(x + 1)} \leq \dfrac{1}{x\ln(x+1)}
\end{equation*}$.
After that you can simply substitute $x+1=t$ and figure out that $\displaystyle \int_{1}^{\infty} \dfrac{1}{(x+1)\ln(x+1)} \ dx$ diverges!
Now it's simple to figure out that $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)}  \ dx$ diverges from the comparison test
