Evaluating $\int_{0}^{1}\frac{x-1}{(x+1)\ln x} dx $ Evaluate the following integral $$\displaystyle I=\int_{0}^{1}\frac{x-1}{(x+1)(\ln x)} \mathrm{d}x $$
My work: I tried it by letting $\displaystyle I(a)=\int_{0}^{1}\frac{(x-1)x^a}{(x+1)(\ln x)} \mathrm{d}x$ and then $\displaystyle I'(a)=\int_{0}^{1}\frac{(x-1)x^a}{x+1} \mathrm{d}x$.
Now $\displaystyle I'(a)=\int_{0}^{1}x^a \mathrm{d}x-\int_{0}^{1}\frac{2x^a}{x+1} \mathrm{d}x$
Now if $\displaystyle J(a)=\int_{0}^{1}\frac{x^a}{x+1}\mathrm{d}x$ , then by applying integration by parts, we get the reccurence relation $J(a)+J(a-1)=\dfrac{1}{a}$ and we can solve it then, but the thing is, we neet to find $I(0)$ ,so even if we compute $J(a)$, it wouldn't be defined at $0$ and so would $I(a)$, then how do I find $I'(a)$ by other method?
I also tried the substitution $x \to \frac{1}{x}$, which yields $\displaystyle I=\int_{1}^{\infty}\frac{(x-1)}{(x+1)(\ln x)} \mathrm{d}x$ and when I saw their graphs, it clearly doesn't seem that the area under the graph of this function from $0$ to $1$ and from $1$ to $\infty$ are equal.
I would appreciate if someone could  continue from my method and other solutions are also welcomed...

Here, we will more focus on answering to OP's specific questions:

1. The substitution $x\mapsto1/x$ yields
$$ I = \int_{1}^{\infty} \frac{x-1}{x^2(x+1)\log x} \, \mathrm{d}x.
$$
So it seems that OP made a mistake when applying the substitution.

2. Continuing from OP's approach, the recurrence relation and $J(\infty)=0$ together imply
$$ J(a)
= \frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+3}-\dots,$$
and hence
\begin{align*}
-I'(a)
&= -\frac{1}{a+1} + 2 \biggl( \frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+3}-\dots \biggr) \\
&= \biggl( \frac{1}{a+1} - \frac{2}{a+2} + \frac{1}{a+3} \biggr) + \biggl( \frac{1}{a+3} - \frac{2}{a+4} + \frac{1}{a+5} \biggr) + \dots \\
&= \sum_{n=1}^{\infty} \biggl( \frac{1}{a+2n-1} - \frac{2}{a+2n} + \frac{1}{a+2n+1} \biggr).
\end{align*}
Now we integrate both sides from $0$ to $\infty$. Then the left-hand side becomes $I(0)$ by $I(\infty) = 0$. On the other hand, each term of the summation is non-negative, so we can apply the Fubini-Tonelli Theorem to interchange the order of integration and summation to get
\begin{align*}
I(0)
&= \int_{0}^{\infty} (-I'(a)) \, \mathrm{d}a \\
&= \sum_{n=1}^{\infty} \int_{0}^{\infty} \biggl( \frac{1}{a+2n-1} - \frac{2}{a+2n} + \frac{1}{a+2n+1} \biggr) \, \mathrm{d}a \\
&= \sum_{n=1}^{\infty} (-\log(2n-1) + 2 \log(2n) - \log(2n+1)) \\
&= \log \Biggl( \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \Biggr).
\end{align*}
By using the Wallis product formula, the product term reduces to $\frac{\pi}{2}$, proving
$$ I(0) = \log\left(\frac{\pi}{2}\right). $$
