I believe that by integration by parts for $\int_0^1u\cdot dv$, see the dot here to define the formula (after we sum for a sufficently large integer) $$\sum_{k=3}^\infty\int_0^1\frac{d}{dx}x^{\mu(k)}\cdot\frac{x^{k-1}}{k^2}dx=\frac{1}{\zeta(3)}-(1-\frac{1}{2^3})-2\sum_{\substack{k\geq 3},\mu(k)=-1}\frac{1}{k^3(k-2)},$$ where $\mu(k)$ is the Möbius function.
And in the other hand with a new integration by parts defined as $\int_0^1u\cdot dv$, see the dot here to define the formula $$\sum_{k=3}^\infty\int_0^1\frac{x^{k-1}}{k^2}\cdot\frac{d}{dx}x^{\mu(k)}dx= (\frac{\pi^2}{6}-1-\frac{1}{4})-\sum_{\substack{k\geq 3},\mu(k)=0}\frac{1}{k^2}$$
$$\qquad\qquad\qquad\qquad-\sum_{\substack{k\geq 3},\mu(k)=1}\frac{k+1}{k^3}-\sum_{\substack{k\geq 3},\mu(k)=-1}\frac{k-1}{k^2(k-2)}.$$ Then notice that if there are no mistakes it is possible to get $\frac{1}{\zeta(3)}$ as function of $\sum_{\substack{k\geq 3},\mu(k)=0}\frac{1}{k^2}$, since the term $$-\left(\sum_{\substack{k\geq 3},\mu(k)=1}\frac{k+1}{k^3}-\sum_{\substack{k\geq 3},\mu(k)=-1}\frac{k+1}{k^3}\right)=$$ is equals to $$-\left(\frac{1}{\zeta(2)}-(1-\frac{1}{2^2})+\frac{1}{\zeta(3)}-(1-\frac{1}{2^3})\right).$$
Question. Please can you find where was the first of my mistakes? I say this because the computations were tedious. If it is possible/feasible, can you obtain an identity with mathematical meaning or the final identity that I've presumed (I say if you can get a simple identity involving the $\frac{1}{\zeta(3)}$ from this approach)? Thanks in advance.
I've tried different integrands in integrals as $$\int_0^1\frac{d}{dx}x^{\mu(k)}\cdot d\left(\frac{x^{k-1}}{k^2}\right),$$ defined also for $k$ sufficiently large (or leaving easy calculations for the corresponding tail), say us $k\geq 4$.