Closed form of the integral $$I=\int_{-1}^1 \frac{\sin\left(\frac{\sinh x}{x}\right)\cdot\log\left(\frac{1+x}{1+x^2}\right)}{x} \space\text{d}x$$

According to Wolfram Alpha, the integral comes out to $$I=2.1607...$$

I don't have a clue on how to solve this because of how the $x$ value is deeply nested here. An answer I am looking for does the following:

a) Showing each step. All of your steps also much use only the real plane. I.E. Real Analysis

b) You don't have to state theorems in your work, but if what you did is not obvious, just explain.

Thanks for any help.

EDIT: this integral has a specific application. That application is that I am just beginning to work on nested integral arguments. The further I get in my study, I consistently give myself challenge problems. Strangely, after a while of trying to crack this, I got nowhere! I just wanted to see how to do this and where my mistake lay. (On original trial of the problem, I tried to substitute $x^2$ with $-u^2$ and ignored the fact that $x \not\to -u$)
I will give some of my own estimates.. Note that

$$
\begin{align*}
\int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log\left(\frac{1+x}{1+x^2}\right) \mathrm{d}x
&amp; = \int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \Bigl[ \log\left( 1 + x\right) -  \log\left(1 + x^2\right) \Big] \mathrm{d}x \\
&amp; = \int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log(1+x) \mathrm{d}x
\end{align*}
$$
Due to the symmetry of the last integral. For example split it at $x=0$ and use the substitution $x \mapsto -x$ on the integral over $[-1,0]$. Again we can map the integral onto $[0,1]$. Introducing the function
$$
f(x) = \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log(1 + x)
$$ 
The integral can then be written as
$$
\begin{align*}
\int_{-1}^{1} f(x) \,\mathrm{d}x 
&amp; = \int_{0}^{1} f(x) + f(-x)\,\mathrm{d}x \\
&amp; = \int_0^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log\left(\frac{1+x}{1-x}\right)\,\mathrm{d}x
\end{align*}
$$
The reason for these algebraic manipulations is that the taylor expansion of the last expression converges much faster. It only contains $\sin x$ and $\log(1 \pm x)$ which are both very well known. We have

$$
\log \frac{1+x}{1-x} = 2\sum_{k=0}^\infty \frac{x^{1+2k}}{1+2k}\\
\frac{\sinh x}{x} = \sum_{j=0}^\infty \frac{x^{2j}}{(1+2j)!}
$$
Combining these and finding the taylor expansion (allowed since $|x|\leq 1$ we get 
$$
f(x) \sim 2 \sin(1) + \frac{x^2}{3} \bigl(2 \sin(1) + \cos(1)\bigr)+ \frac{x^4}{180} \bigl(67 \sin(1) + 23 \cos(1)\bigr)
$$
Integrating from $0$ to $1$ we get
$$
\int_{0}^{1} f(x) \,\mathrm{d}x = \frac{689}{300} \sin(1)+\frac{41}{300}\cos(1) \approx 2.006419677
$$
Which is - if I dare so myself - a darn good estimate. Seeing from the taylor expansion that the answer will always be on the form $a \sin 1 + b \cos 1$, one can improve these constants somewhat. For example 
$$
\frac{268}{125} \sin 1 + \frac{33}{50} \cos 1
$$
Which is good for 9 decimals give or take.
I will give some of my own estimates.. Note that

$$
\begin{align*}
\int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log\left(\frac{1+x}{1+x^2}\right) \mathrm{d}x
&amp; = \int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \Bigl[ \log\left( 1 + x\right) -  \log\left(1 + x^2\right) \Big] \mathrm{d}x \\
&amp; = \int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log(1+x) \mathrm{d}x
\end{align*}
$$
Due to the symmetry of the last integral. For example split it at $x=0$ and use the substitution $x \mapsto -x$ on the integral over $[-1,0]$. Again we can map the integral onto $[0,1]$. Introducing the function
$$
f(x) = \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log(1 + x)
$$ 
The integral can then be written as
$$
\begin{align*}
\int_{-1}^{1} f(x) \,\mathrm{d}x 
&amp; = \int_{0}^{1} f(x) + f(-x)\,\mathrm{d}x \\
&amp; = \int_0^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log\left(\frac{1+x}{1-x}\right)\,\mathrm{d}x
\end{align*}
$$
The reason for these algebraic manipulations is that the taylor expansion of the last expression converges much faster. It only contains $\sin x$ and $\log(1 \pm x)$ which are both very well known. We have

$$
\log \frac{1+x}{1-x} = 2\sum_{k=0}^\infty \frac{x^{1+2k}}{1+2k}\\
\frac{\sinh x}{x} = \sum_{j=0}^\infty \frac{x^{2j}}{(1+2j)!}
$$
Combining these and finding the taylor expansion (allowed since $|x|\leq 1$ we get 
$$
f(x) \sim 2 \sin(1) + \frac{x^2}{3} \bigl(2 \sin(1) + \cos(1)\bigr)+ \frac{x^4}{180} \bigl(67 \sin(1) + 23 \cos(1)\bigr)
$$
Integrating from $0$ to $1$ we get
$$
\int_{0}^{1} f(x) \,\mathrm{d}x = \frac{689}{300} \sin(1)+\frac{41}{300}\cos(1) \approx 2.006419677
$$
Which is - if I dare so myself - a darn good estimate. Seeing from the taylor expansion that the answer will always be on the form $a \sin 1 + b \cos 1$, one can improve these constants somewhat. For example 
$$
\frac{268}{125} \sin 1 + \frac{33}{50} \cos 1
$$
Which is good for 9 decimals give or take.