<s>[INST] Evaluate $\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt$ when $N=3$ or $N=5$ Let the Cauchy product $$(\zeta(3))^2=\sum_{n=1}^\infty c_n,$$
where $$c_n=\sum_{k=1}^n\frac{1}{k^3(n-k+1)^3},$$ 
and $\zeta(3)$ is the Apèry constant.
Taking $f(x)=\frac{1}{x^3(N-x+1)^3}$ in Abel's identity after tedious computations using the division of polynomials, since $f'(x)=\frac{-3N+6x-3}{x^4(N-x+1)^4}$ we've 
$$\sum_{n\leq N}\frac{1}{n^3(N-n+1)^3}=\frac{1}{N^2}-\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt+O \left( \frac{1}{N^6} \right). $$
Then I understand that I can't get a closed form from this way since I have an error term that I don't know how compute it, but I've asked to me if at least we can compute previous integral, seems a hard integral, but Wolfram Alpha knows how compute particular values of previous integral. Then since I would like learn more mathematics 

Question. Can you explain how get a closed-form, for example for $N=3$ or $N=5$ by specializacion of the previous integral 
  $$\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt?$$ 

It is neccesary only one case $N=3$ or $5$, I say these because I see that these are posible with this online tool. I will accept one answer with good details. Very thanks much!
 [/INST] $$\frac{1}{n(N-n)} = \frac{1}{N}\left(\frac{1}{n}+\frac{1}{N-n}\right) \tag{1}$$
gives, by squaring:
$$\frac{1}{n^2(N-n)^2} = \frac{1}{N^2}\left(\frac{1}{n^2}+\frac{1}{(N-n)^2}+\frac{\frac{2}{N}}{n}+\frac{\frac{2}{N}}{N-n}\right)\tag{2} $$
as well as:
$$ \frac{1}{n^3 (N-n)^3} = \frac{1}{N^3}\left(\frac{1}{n^3}+\frac{1}{(N-n)^3}+\frac{3N}{n^2(N-n)^2}\right)\tag{3}$$
hence it follows that:
$$ \sum_{k=1}^{n}\frac{1}{k^3(n+1-k)^3} = \frac{2 H_n^{(3)}}{n^3}+\frac{6 H_n^{(2)}}{n^4}+\frac{12 H_n}{n^5}\tag{4} $$
and the same strategy (partial fraction decomposition) can be applied to the integral resulting from Abel's identity. However, Abel's identity is not really needed to compute the LHS of $(4)$.
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