Let's denote the integral in question as
$$I=\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}dx.\tag1$$
Changing the variable $x=y^{1/10}$ and renaming $y$ back to $x$ we get
$$I=\int_0^1\frac{x^{2/5}+x^{1/5}-x^{1/10}-1-\ln\sqrt x}{(x-1)\ln x}dx.\tag2$$
Some elementary transformations show that
$$I=\mathcal{J}(2/5)+\mathcal{J}(1/5)-\mathcal{J}(1/10),\tag3$$
where we introduced notation
$$\mathcal{J}(q)=\int_0^1\frac{x^q-1-q\,\ln x}{(x-1)\ln x}dx.\tag4$$
The integral $\mathcal{J}(q)$ can be evaluated as follows:
$$\begin{align}\mathcal{J}(q)=\int_0^1\int_0^q\frac{x^p-1}{x-1}dp\,dx=\int_0^q\underbrace{\int_0^1\frac{x^p-1}{x-1}dx}_{\text{DLMF 5.9.16}}\,dp\\=\int_0^q H_p\,dp=q\cdot\gamma+\ln\Gamma(q+1),\end{align}\tag5$$
where $H_p$ are harmonic numbers: $H_p$$\,=\,$$\gamma$$\,+\,$$\psi_0$$(p+1)$, and $\psi_0$ is the digamma function: $\psi_0(x)=\frac{d}{dx}\ln\,$$\Gamma$$(x)$. Let me mention that the formula DLMF 5.9.16 becomes particularly obvious for positive integer $p$, when $H_p=\sum_{n=1}^pn^{-1}$.
Pluging $(5)$ back into $(3)$, we get
$$I=\frac12\gamma+\ln\frac45+\ln\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}.\tag6$$
From the formula $(74)$ on this MathWorld page we know that
$$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}=\frac{\sqrt[5]2\,\sqrt\pi}{\sqrt[4]5\,\sqrt\phi}.\tag7$$
(see the paper Raimundas Vidūnas, Expressions for values of the gamma function for a proof).
Making use of this formula, we get the final result
$$I=\frac12\gamma+\frac{11}5\ln2-\frac54\ln5+\frac12\ln\pi-\frac12\ln\phi.\tag8$$
First, we need a preliminary result:
First preliminary result :
$$\ln\left(2 \sinh\left(\frac{x}{2}\right)\right)=\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \tag{1}$$
Proof:
$$
\begin{aligned}
\ln\left( \sinh(x)\right)&=\ln\left( \frac{1}{2} \left( e^{x}-e^{-x} \right)\right)\\
&=-\ln 2+\ln\left( e^{x}-e^{-x} \right)\\
&=-\ln 2+\ln\left( \frac{e^{-x}}{e^{-x}} \left( e^{x}-e^{-x} \right)\right)\\
&=-\ln 2+x+\ln\left( 1-e^{-2x} \right)\\
&=-\ln 2+x-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n} \qquad \blacksquare
\end{aligned}
$$
Letting $x \to \frac{x}{2}$ completes the proof
Than:
$$
\begin{aligned}
\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}&=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx \\
&=-2\int_0^{2\ln(\phi)}x\left(\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \right)\,dx & \left( \text{by eq. (1)}\right)\\
&=-\int_0^{2\ln(\phi)}x^2\,dx+2\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{2\ln(\phi)} xe^{-kx}\,dx\\
&=-\frac{8}{3}\ln^3(\phi)+2\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{2\ln(\phi)\phi^{-2k}}{k}+\frac{1}{k}\int_0^{2\ln(\phi)} e^{-kx}\,dx \right)\\
&=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^2}+2\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{k^2}-\frac{(\phi^{-2})^k}{k^2}\right)\\
&=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})+2\zeta(3)-2\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^3}\\
&=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})-2\operatorname{Li}_3(\phi^{-2})+2\zeta(3)\\
&=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\left( \frac{\pi^{2}}{15}-\ln ^{2} \phi\right)-2\left(\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15} \right)+2\zeta(3)\\
&=-\frac{8}{3}\ln^3(\phi)+4\ln^3(\phi)-\frac{4}{3}\ln^3(\phi)-\frac85\zeta(3)+2\zeta(3)\\
&=\frac25\zeta(3) \qquad \blacksquare
\end{aligned}
$$
Which is the representation of $\zeta(3)$ that Apery used to prove the irrationality of $\zeta(3)$.
Note that we used
$\mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) =\frac{\pi^{2}}{15}-\ln ^{2} \phi$
A proof can be found here in my blog
And:
$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$
Proof of the second relation:
Recall the Trilogarithm identity proved here
$
\operatorname{Li}_{3}(x)+\operatorname{Li}_{3}(1-x)+\operatorname{Li}_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}$
And the polylogarithm relation:
$\operatorname{Li}_{n}(x)+\operatorname{Li}_{n}(-x)=2^{1-n}\operatorname{Li}_{n}(x^2)$
Example, letting $x=\phi^{-1}$ in the last relation we obtain
$$\operatorname{Li}_{n}(\phi^{-1})+\operatorname{Li}_{n}(-\phi^{-1})=\frac14\operatorname{Li}_{n}(\phi^{-2})$$
Claim:
$$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$$
Proof:
If we let $x=\phi^{-1}$ in $(2)$ we obtain
$$
\begin{aligned}
&\operatorname{Li}_{3}\left(\phi^{-1}\right)+\operatorname{Li}_{3}(1-\phi^{-1})+\operatorname{Li}_{3}\left(1-\frac{1}{\phi^{-1}}\right)=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}-\frac{\ln ^{2}(\phi) \ln (1-\phi^{-1})}{2}\\
&\operatorname{Li}_3\left(\phi^{-1} \right)+\operatorname{Li}_3\left(\phi^{-2} \right)+\operatorname{Li}_3\left(-\phi^{-1} \right)=\zeta(3)-\frac{\ln^3(\phi)}{6}-\frac{\pi^2\ln(\phi)}{6}-\frac{\ln^2(\phi)\ln(\phi^{-2})}{2}\\
&\frac14\operatorname{Li}_{3}\left(\phi^{-2}\right)+\operatorname{Li}_{3}(\phi^{-2})=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}+\ln ^{3}(\phi) \\
&\frac54\operatorname{Li}_{3}\left(\phi^{-2}\right)=\zeta(3)+\frac{5\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6} \\
&\operatorname{Li}_{3}\left(\phi^{-2}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2}\ln (\phi)}{15} \qquad \blacksquare \\
\end{aligned}
$$
Best Answer
As said in my first comment above, the integral is divergent (so: wolframaplha is wrong). Indeed, let $a_k:=\frac\pi2+2k\pi$ with $\Bbb N\ni k\to+\infty$ then $$\int_{\sqrt{a_k-\frac\pi4}}^{\sqrt{a_k+\frac\pi4}}e^{x^2}\frac{\sin(x^2)}{x^2}dx=\frac12\int_{a_k-\frac\pi4}^{a_k+\frac\pi4}e^y\frac{\sin y}{y^{3/2}}dy $$ $$\ge\frac{e^{a_k-\frac\pi4}}{2\sqrt2(a_k+\frac\pi4)^{3/2}}\to+\infty$$