calculus - How to show $int_0^1 left( frac{1}{ln(1+x)} - frac{1}{x} right) ,dx$ converges?

I need to show that $$\int_0^1 \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) \,dx$$ converges, given that $$\lim_{x\rightarrow0^+} \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) = \frac{1}{2}$$

I'm not sure how to do this because my text gave a very brief treatment on applying the limit comparison test for the improper integral of the second kind. Here is what I've tried:

I define the test function to be $g(t)=1$. Therefore,

$$\frac{\left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right)}{g(t)} = \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) \tag{1}$$

So, $(1)$ tends to $\frac{1}{2}$ as $t$ tends to $0^+$ and $$\int_0^1 g(t) \,dt$$ is convergent. Therefore, by the limit comparison test, the given integral is convergent.

Did I do the test correctly?

EDIT

My text's solution (which I find to be vague) is:

What does 'repair' the integrand mean?

Answer

I'll give a more general answer, which should explain what “repairing the function” means.

Suppose $f$ is a continuous function defined on the interval $(a,b]$ (where $a$$

\lim_{x\to a^+}f(x)=r
$$
is finite.

Define $g(x)=f(x)$ for $a\begin{align}
F(x)&=\int_{x}^b f(t)\,dt &&\text{for $x\in(a,b]$}\\
G(x)&=\int_{x}^b g(t)\,dt &&\text{for $x\in[a,b]$}
\end{align}
The two functions $F$ and $G$ are continuous (even differentiable, by the fundamental theorem of calculus) and coincide on $(a,b]$, so

$$
\lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=G(a)
$$